Class 12 Maths Solutions For Matrices
Matrix A rectangular array of mn numbers in the form of m horizontal lines (called rows) and n vertical lines (called columns) is called a matrix of order m by n, written as an m x n matrix.
Such an array is enclosed by [ ] or ( ).
Each of the mn numbers constituting the matrix is called an element or an entry of the matrix.
Usually, we denote a matrix by a capital letter.
The plural of matrix is matrices.
Examples (1) A = \(\left[\begin{array}{rrr}
3 & 5 & -4 \\
0 & 1 & 9
\end{array}\right]\) is a matrix, having 2 rows and 3 columns.
Its order is 2 x 3 and it has 6 elements.
(2) B = \(\left[\begin{array}{rrrr}
9 & 4 & \sqrt{2} & -1 \\
1 & 8 & -3 & 2 \\
6 & 0 & 5 & 7
\end{array}\right]\) is a matrix, having 3 rows and 4 columns. Its order is 3 x 4 and it has 12 elements.
How to Describe a Matrix
To locate the position of a particular element of a matrix, we have to specify the number of the row and that of the column in which the element occurs.
An element occuring in the ith row and jth column of a matrix A will be called the (i,j)th element of A, to be denoted by air.
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In general, as m x n matrix A may be written as
A = \(\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & \ldots & a_{1 n} \\
a_{21} & a_{22} & a_{23} & \ldots & a_{2 n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{i 1} & a_{i 2} & a_{i 3} & \ldots & a_{i n} \\
\ldots & \ldots & \ldots & \ldots & \ldots \\
a_{m 1} & a_{m 2} & a_{m 3} & \cdots & a_{n n}
\end{array}\right]=\left[a_{i j}\right]_{m \times n} .\)
Example 1 Consider the matrix A = \(\left[\begin{array}{rrr}
3 & -2 & 5 \\
6 & 9 & 1
\end{array}\right]\)
Clearly, the element in the 1st row and 2nd column is -2.
So, we write a12 = -2.
Similarly, a11 = 3; a12 = -2; a13 = 5; a21 = 6; a22 = 9 and a23 = 1.
Example 2 Construct a 3 x 2 matrix whose elements are given by aij = (i+2j).
Solution
A 3 x 2 matrix has 3 rows and 2 columns.
In general, a 3 x 2 matrix is given by
A = \(\left[\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right]_{3 \times 2}\)
Thus aij = (i+2j) for i=1, 2,3 and j = 1,2.
∴ a11 = (1+2×1) = 3; a12 = (1+2×2) = 5;
a21 = (2+2×1) = 4; a22=(2+2×2) = 6;
a31 = (3+2×1) = 5; a32 = (3+2×2) = 7.
Hence, A = \(\left[\begin{array}{ll}
3 & 5 \\
4 & 6 \\
5 & 7
\end{array}\right]_{3 \times 2}\)
Example 3 Construct a 2 x 3 matrix whose elements are given by aij = \(\frac{1}{2}|5 i-3 j| .\)
Solution
A 2 x 3 matrix has 2 rows and 3 columns.
In general, a 2 x 3 matrix is given by
A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]_{2 \times 3}\)
Thus, \(a_{i j}=\frac{1}{2}|5 i-3 j|\), where i = 1,2 and j = 1,2,3.
∴ \(a_{11}=\frac{1}{2}|5 \times 1-3 \times 1|=\frac{1}{2} \cdot|2|=\frac{1}{2} \times 2=1 ;\)
\(a_{12}=\frac{1}{2}|5 \times 1-3 \times 2|=\frac{1}{2} \cdot|5-6|=\frac{1}{2} \cdot|-1|=\frac{1}{2} \times 1=\frac{1}{2} \text {; }\) \(a_{13}=\frac{1}{2}|5 \times 1-3 \times 3|=\frac{1}{2} \cdot|5-9|=\frac{1}{2} \cdot|-4|=\frac{1}{2} \times 4=2 \text {; }\) \(a_{21}=\frac{1}{2}|5 \times 2-3 \times 1|=\frac{1}{2} \cdot|10-3|=\frac{1}{2} \cdot|7|=\frac{1}{2} \times 7=\frac{7}{2} ;\) \(a_{22}=\frac{1}{2}|5 \times 2-3 \times 2|=\frac{1}{2} \cdot|10-6|=\frac{1}{2} \cdot|4|=\frac{1}{2} \times 4=2\) \(a_{23}=\frac{1}{2}|5 \times 2-3 \times 3|=\frac{1}{2} \cdot|10-9|=\frac{1}{2} \cdot|1|=\frac{1}{2} \times 1=\frac{1}{2}\)Hence, A = \(\left[\begin{array}{lll}
1 & \frac{1}{2} & 2 \\
\frac{7}{2} & 2 & \frac{1}{2}
\end{array}\right]\).
Example 4 If a matrix has 12 elements, what are the possible orders it can have?
Solution
We know that a matrix of order m x n has mn elements.
Hence, all possible orders of a matrix having 12 elements are (12×1), (1×12), (6×2), (2×6), (4×3) and (3×4).
Various Types of Matrices
Row Matrix A matrix having only one row is known as a row matrix or a row vector.
Examples (1) A = [5 18] is a row matrix of order 1 x 2.
(2) B = [2 √5 -9 0] is a row matrix of order 1 x 4.
Column Matrix A matrix having only one column is known as a column matrix or a column vector.
Examples (1) A = \(\left[\begin{array}{r}
2 \\
7 \\
-3
\end{array}\right]\) is a column matrix of order 3 x 1.
(2) B = \(\left[\begin{array}{l}
6 \\
4
\end{array}\right]\) is a column matrix of order 2 x 1.
Zero Or Null Matrix A matrix each of whose elements is zero is called a zero matrix or a null matrix.
Example The matrices [0], [0 0], \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\) and \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) are null matrices of order (1×1), (1×2), (2×2) and (2×3) respectively.
Square Matrix A matrix having the same number of rows and columns is called a square matrix.
A matrix of order (n x n) is called a square matrix of order n or an n-rowed square matrix.
A matrix of order m x n, where m ≠ n, is called a rectangular matrix.
Examples (1) The matrix \(\left[\begin{array}{rr}
3 & 2 \\
6 & -5
\end{array}\right]\) is a 2-rowed square matrix.
(2) The matrix \(\left[\begin{array}{ccc}
5 & 3 & 6 \\
7 & \sqrt{2} & -4 \\
-9 & \frac{1}{3} & 0
\end{array}\right]\) is a 3-rowed square matrix.
Diagonal Elements Of A Matrix Let A = [aij]mxn be an m x n matrix. Then, the elements aij for which i = j, are called the diagonal elements of A.
Thus, the diagonal elements of A = [aij]mxn are a11, a22, a33, a44, etc.
The line along which the diagonal elements lie is called the diagonal of the matrix.
Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{5} & \frac{5}{8} & 7 \\
6 & -4 & \sqrt{2}
\end{array}\right]\) Then, the diagonal elements of A are: a11 = 3, a22 = \(\frac{5}{8}\), a33 = √2.
Diagonal Matrix A square matrix in which every nondiagonal element is zero is called a diagonal matrix.
If A = [aij]mxn be a diagonal matrix then aij=0 when i ≠ j and we write it as
A = diag[a11, a22, a33,…, ann].
Example Let A = \(\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & -2
\end{array}\right]\). Then, A is a diagonal matrix. We may write it as, A = diag[6,4,-2].
Scalar Matrix A square matrix in which every nondiagonal element is zero and all diagonal elements re equal is known as a scalar matrix.
Examples (1) A = \(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\) is a scalar matrix of order 2.
(2) B = \(\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -3
\end{array}\right]\) is a scalar matrix of order 3.
Unit Matrix A square matrix in which every nondiagonal element is 0 and every diagonal element is 1 is called a unit matrix or an identity matrix.
Thus, a square matrix [aij]nxn is a unit matrix if
\(a_{i j}=\left\{\begin{array}{l}0 \text { when } i \neq j, \\
1 \text { when } i=j .
\end{array}\right.\)
A unit matrix of order n will be denoted by In or simply by I.
Examples (1) I2 = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) is a unit matrix of order 2.
(2) I3 = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) is a unit matrix of order 3.
Comparable Matrices Two matrices A and B are said to be comparable if they are of the same order, i.e., they have the same number of rows and the same umber of columns.
Example A = \(\left[\begin{array}{rrr}
2 & -5 & 1 \\
0 & 3 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
3 & 7 & 0 \\
1 & 4 & -9
\end{array}\right]\) are comparable matrices, each being of order (2×3).
Equal Matrices Two matrices A and B are said to be equal, written as A = B, if they are of the same order and their corresponding elements are equal.
Example 1 Find x, y, z when \(\left[\begin{array}{ll}
5 & 3 \\
x & 7
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 7
\end{array}\right]\)
Solution
Since the corresponding elements of equal matrices are equal, we have
\(\left[\begin{array}{ll}
5 & 3 \\
x & 7
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 7
\end{array}\right]\) ⇔ x = 1, y = 5 and z = 3.
Example 2 Find x, y, z w when \(\left[\begin{array}{ll}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\).
Solution
We know that in equal matrices, the corresponding elements are equal.
∴ \(\left[\begin{array}{rr}
x-y & 2 x+z \\
2 x-y & 3 z+w
\end{array}\right]=\left[\begin{array}{rr}
-1 & 5 \\
0 & 13
\end{array}\right]\)
⇔ x – y = -1, 2x – y = 0, 2x + z = 5 and 3z + w = 13.
Solving the first two equations, we get x = 1 and y = 2.
Putting x = 1 in 2x + z = 5, we get z = 3.
Putting z = 3 in 3z + w = 13, we get w = 4.
∴ x = 1, y = 2, z = 3 and w = 4.
Example 3 \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \neq\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right] \text {. }\) Why?
Solution
Since the given null matrices are not comparable, they are not equal.
Operations On Matrices
Mainly we have four operations on matrices, namely:
Transposition, Matrix Addition, Matrix multiplication and Scalar multiplication.
Out of these operations, transposition is a unary operation while matrix addition and matrix multiplication are both binary operations and scalar multiplication is an external composition.
Transposition
Transpose Of A Matrix Let A be an (mxn) matrix. Then, the matrix obtained by interchanging the rows and columns of A is called the transpose of A, denoted by A’ or A’.
If A = [aij]mxn, then A’ = [aij]nxm.
Remarks (1) If A is an (mxn) matrix, then A’ is an (nxm) matrix.
(2) (i,j)th element of A = (j,i)th element of A’.
Examples (1) If A = \(\left[\begin{array}{rrr}
2 & 3 & -1 \\
4 & -2 & 5
\end{array}\right]\), then \(A^t=\left[\begin{array}{rr}
2 & 4 \\
3 & -2 \\
-1 & 5
\end{array}\right]\)
Here A is a (2×3) matrix and A’ is a (3×2) matrix.
(2) If B = \(\left[\begin{array}{r}
3 \\
-4 \\
6
\end{array}\right]\), then Bt = [3 -4 6].
Here, B is a (3×1) matrix and Bt is a (1×3) matrix.
Theorem (Involution) For any matrix A, prove that (A’)’ = A.
Proof
Let A = [aij]mxn matrix ⇒ A’ is an (nxm) matrix.
⇒ (A’)’ is a (mxn) matrix.
∴ A and (A’)’ are matrices of the same order.
Also, (i,j)th element of A = (j,i)th element of A’
= (i,j)th element of (A’)’.
Thus, A and (A’)’ are comparable matrices having their corresponding elements equal.
Hence, (A’)’ = A.
Symmetric Matrix A square matrix A is said to be symmetric if A’ = A.
Thus, A is symmetric ⇔ aji = aij.
Examples (1) If A = \(\left[\begin{array}{rr}
4 & 2 \\
2 & -3
\end{array}\right]\), then A’ = \(\left[\begin{array}{rr}
4 & 2 \\
2 & -3
\end{array}\right]\) = A.
∴ A is symmetric.
(2) If B = \(\left[\begin{array}{rrr}
6 & 8 & -4 \\
8 & 3 & 0 \\
-4 & 0 & 5
\end{array}\right]\), then B’ = \(\left[\begin{array}{rrr}
6 & 8 & -4 \\
8 & 3 & 0 \\
-4 & 0 & 5
\end{array}\right]\) = B.
∴ B is symmetric.
Skew-Symmetric Matrix A square matrix A is said to be skew-symmetric if A’ = -A, where -A is the matrix obtained by replacing each element of A by its negative.
Remark A is skew-symmetric ⇒ A’ = -A
⇒ aji = -aij, where A = [aij]nxn
⇒ aii = -aii
⇒ 2aii = 0 ⇒ aii = 0
⇒ every diagonal element of A is 0.
Thus, every diagonal element of a skew-symmetric matrix is 0.
Examples (1) If A = \(\left[\begin{array}{rr}
0 & 8 \\
-8 & 0
\end{array}\right]\), then A’ = \(\left[\begin{array}{rr}
0 & -8 \\
8 & 0
\end{array}\right]\) = -A.
Hence, A is skew-symmetric.
(2) Let B = \(\left[\begin{array}{ccc}
0 & h & -g \\
-h & 0 & f \\
-g & f & 0
\end{array}\right]\), then B’ = \(\left[\begin{array}{ccc}
0 & -h & -g \\
h & 0 & -f \\
g & f & 0
\end{array}\right]\) = -B.
Hence, B is skew-symmetric.
Addition Of Matrices
Let A and B be two comparable matrices, each of order (mxn). Then, their sum (A+B) is a matrix of order (mxn), obtained by adding the corresponding elements of A and B.
Example 1 If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 4 & 5 \\
1 & 2 & 3
\end{array}\right]\) then A and B are matrices order 2 x 2 and 2 x 3 respectively.
So, A and B are not comparable.
Hence, A+B is not defined.
Example 2 Let A = \(\left[\begin{array}{lll}
5 & 0 & -2 \\
3 & 2 & -7
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
4 & -3 & -6 \\
-1 & 0 & 4
\end{array}\right] \text {. }\)
Clearly, each one of A and B is a 2 x 3 matrix.
∴ A + B is defined.
We have: A+B = \(\left[\begin{array}{lll}
5+4 & 0+(-3) & -2+(-6) \\
3+(-1) & 2+0 & -7+4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
9 & -3 & -8 \\
2 & 2 & -3
\end{array}\right]\)
Some Results on Addition of Matrices
Theorem 1 Matrix addition is commutative, i.e., A + B = B + A for all comparable matrices A and B.
Proof
Let A = [aij]mxn abd B = [bij]mxn. Then,
A + B = [aij]mxn + [bij]mxn
= [aij + bij]mxn [by the definition of addition of matrices]
= [bij + aij]mxn [∵ addition of numbers is commutative]
= [bij]mxn + [aij]mxn = B + A.
Hence, A + B = B + A.
Theorem 2 Matrix addition is associative, i.e., (A + B)+C = A+(B+C) for all comparable matrices A, B and C.
Proof
Let A = [aij]mxn, B = [bij]mxn and C = [cij]mxn. Then,
(A+B)+c = ([aij]mxn + [bij]mxn) + [cij]mxn
= [aij + bij]mxn + [cij]mxn
= [(aij+bij)+cij]mxn
= [aij + (bij + cij)]mxn
[∵ addition of numbers is associative]
= [aij]mxn + [bij+cij]mxn
= [aij]mxn + ([bij]m+n + [cij]mxn) = A + (B+C).
Hence, (A+B)+C = A+(B+C).
Theorem 3 If A is an mxn matrix and O is an mxn null matrix, then A + O = O + A = A.
Proof
Let A = [aij]mxn and O = [bij]mxn,
where bij = 0 for all suffixes i and j.
Then, A + O = [aij]mxn + [bij]mxn = [aij+bij]mxn
= [aij+0]mxn [∵ bij = 0]
= [aij]mxn = A.
∴ A + O = A.
Similarly, O + A = A.
Hence, A + O = O + A = A.
Remark The null matrix O of order m x n is the additive identity in the set of all m x n matrices.
Example 3 Let A = \(\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) and O = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\), then verify that A + O = O + A = A.
Solution
Clearly, each one of A and O is a matrix of order (2×3).
So, (A+O) and (O+A) are both defined.
Now, A + O = \(\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]+\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
3+0 & 5+0 & 4+0 \\
1+0 & 2+0 & -3+0
\end{array}\right]+\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) = A.
And, O + A = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]+\left[\begin{array}{rrr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0+3 & 0+5 & 0+4 \\
0+1 & 0+2 & 0+(-3)
\end{array}\right]=\left[\begin{array}{llr}
3 & 5 & 4 \\
1 & 2 & -3
\end{array}\right]\) = A.
Hence, A + O = O + A = A.
Negative Of A Matrix Let A = [aij]mxn. Then, the negative of A is the matrix (-A) = [-aij]mxn, obtained by replacing each element of A with its corresponding additive inverse. (-A) is called the additive inverse of A.
Example 4 If A = \(\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]\), find (-A) and verify that A + (-A) = (-A) + A = 0.
Solution
Clearly, we have
(-A) = \(\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]\)
Now, A + (-A) = \(\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]+\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
3+(-3) & -2+2 & 0+0 \\
-5+5 & 7+(-7) & \sqrt{2}+(-\sqrt{2})
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=O\)
and, (-A) + A = \(\left[\begin{array}{rrr}
-3 & 2 & 0 \\
5 & -7 & -\sqrt{2}
\end{array}\right]+\left[\begin{array}{rrr}
3 & -2 & 0 \\
-5 & 7 & \sqrt{2}
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-3+3 & 2+(-2) & 0+0 \\
5+(-5) & -7+7 & -\sqrt{2}+\sqrt{2}
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Hence, A + (-A) = (-A) + A = O.
Theorem 4 If A and B are two matrices of the same order then prove that (A+B)’ = (A’+B’).
Proof Let A = [aij]mxn and B = [bij]mxn. Then,
A is an (mxn) matrix, B is an (mxn) matrix
⇒ (A+B) is an (mxn) matrix
⇒ (A + B)’ is an (nxm) matrix.
Also, A is an (mxn) matrix and B is an (mxn) matrix
⇒ A’ is an (nxm) matrix and B’ is an (nxm) matrix
⇒ (A’ + B’) is an (nxm) matrix.
Thus, (A+B)’ and (A’+B’) are comparable matrices.
Also, (j,i)th element of (A+B)’
= (i,j)th element of (A+B)
= (i,j)th element of A + (i,j)th element of B
= (j, i)th element of A’ + (j, i)th element of B’
= (j, i)th element of (A’ + B’).
Thus, (A + B)’ and (A’ + B’) are comparable and their corresponding elements are equal.
Hence, (A+B)’ = (A’ + B’).
Solved Examples
Example 1 Let A = \(\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]\) Verify that (A’)’ = A.
Solution
We have A’ = \(\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]^t=\left[\begin{array}{rr}
2 & 0 \\
3 & -4 \\
-5 & 8
\end{array}\right]\)
⇒ \(\left(A^t\right)^t=\left[\begin{array}{rr}
2 & 0 \\
3 & -4 \\
-5 & 8
\end{array}\right]^t=\left[\begin{array}{rrr}
2 & 3 & -5 \\
0 & -4 & 8
\end{array}\right]=A\)
Hence, (A’)’ = A.
Example 2 Let A = \(\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]\). Verify that A + B = B + A.
Solution
Here, A is a 2 x 3 matrix and B is a 2 x 3 matrix. So, A and B are comparable.
Therefore, (A+B) and (B+A) both exist and each is a 2 x 3 matrix.
Now, A + B = \(\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]+\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]\)
= \(\left[\begin{array}{rll}
2+4 & 3+(-2) & 5+3 \\
-1+2 & 0+6 & 4+(-1)
\end{array}\right]=\left[\begin{array}{lll}
6 & 1 & 8 \\
1 & 6 & 3
\end{array}\right] .\)
And, B + A = \(\left[\begin{array}{rrr}
4 & -2 & 3 \\
2 & 6 & -1
\end{array}\right]+\left[\begin{array}{rrr}
2 & 3 & 5 \\
-1 & 0 & 4
\end{array}\right]\)
= \(\left[\begin{array}{lrr}
4+2 & -2+3 & 3+5 \\
2+(-1) & 6+0 & (-1)+4
\end{array}\right]=\left[\begin{array}{lll}
6 & 1 & 8 \\
1 & 6 & 3
\end{array}\right]\)
Hence, A + B = B + A.
Example 3 Let A = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\). Verify that (A + B) + C = A + (B + C).
Solution
Clearly, each one of the matrices A, B, C is a (3×2) matrix. So, (A + B) + C and A + (B + C) are both defined and each one is a 3 x 2 matrix.
Now, (A+B) = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]+\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lr}
1+3 & -2+1 \\
5+0 & 4+2 \\
3+(-3) & 0+5
\end{array}\right]=\left[\begin{array}{rr}
4 & -1 \\
5 & 6 \\
0 & 5
\end{array}\right]\)
∴ (A+B)+C = \(\left[\begin{array}{rr}
4 & -1 \\
5 & 6 \\
0 & 5
\end{array}\right]+\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\)
= \(\left[\begin{array}{lr}
4+4 & -1+3 \\
5+(-2) & 6+2 \\
0+1 & 5+6
\end{array}\right]=\left[\begin{array}{rr}
8 & 2 \\
3 & 8 \\
1 & 11
\end{array}\right]\)
Also, (B + C) = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-3 & 5
\end{array}\right]+\left[\begin{array}{rr}
4 & 3 \\
-2 & 2 \\
1 & 6
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3+4 & 1+3 \\
0+(-2) & 2+2 \\
-3+1 & 5+6
\end{array}\right]=\left[\begin{array}{rr}
7 & 4 \\
-2 & 4 \\
-2 & 11
\end{array}\right] \text {. }\)
∴ A + (B + C) = \(\left[\begin{array}{rr}
1 & -2 \\
5 & 4 \\
3 & 0
\end{array}\right]+\left[\begin{array}{rr}
7 & 4 \\
-2 & 4 \\
-2 & 11
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1+7 & -2+4 \\
5+(-2) & 4+4 \\
3+(-2) & 0+11
\end{array}\right]=\left[\begin{array}{rr}
8 & 2 \\
3 & 8 \\
1 & 11
\end{array}\right] \text {. }\)
Hence, (A + B) + C = A + (B + C).
Example 4 Find the additive inverse of the matrix A = \(\left[\begin{array}{rrr}
2 & -5 & 0 \\
4 & 3 & -1
\end{array}\right]\).
Solution
The additive inverse of the given matrix A is the matrix -A, given by
\(-A=\left[\begin{array}{ccc}-2 & -(-5) & 0 \\
-4 & -3 & -(-1)
\end{array}\right]=\left[\begin{array}{ccc}
-2 & 5 & 0 \\
-4 & -3 & 1
\end{array}\right]\)
Subtraction Of Matrices If A and B are two comparable matrices then we define (A – B) = A + (-B).
Example 5 If A = \(\left[\begin{array}{rrr}
2 & -3 & 1 \\
0 & 7 & -9
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & -3 \\
4 & 8 & -4
\end{array}\right]\), find (A – B).
Solution
We have, (-B) = \(\left[\begin{array}{lll}
-1 & -2 & 3 \\
-4 & -8 & 4
\end{array}\right]\)
∴ (A – B) = A + (-B)
= \(\left[\begin{array}{rrr}
2 & -3 & 1 \\
0 & 7 & -9
\end{array}\right]+\left[\begin{array}{lll}
-1 & -2 & 3 \\
-4 & -8 & 4
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2+(-1) & -3+(-2) & 1+3 \\
0+(-4) & 7+(-8) & -9+4
\end{array}\right]=\left[\begin{array}{rrr}
1 & -5 & 4 \\
-4 & -1 & -5
\end{array}\right]\)
Hence, (A – B) = \(\left[\begin{array}{rrr}
1 & -5 & 4 \\
-4 & -1 & -5
\end{array}\right] \text {. }\)
Example 6 Prove that the sum of two symmetric matrices is symmetric.
Solution
Let A and B be two symmetric matrices of the same order.
Then, A’ = A and B’ = B.
∴ (A + B)’ = A’ + B’ = (A + B) [∵ A’ = A and B’ = B].
Hence, (A + B) is symmetric.
Example 7 Prove that the sum of two skew-symmetric matrices is a skew-symmetric matrix.
Solution
Let A and B be two skew-symmetric matrices. Then,
A’ = -A and B’ = -B.
∴ (A + B)’ = (A’ + B’) = (-A) + (-B) = -(A + B).
Hence, (A + B) is skew-symmetric.
Example 8 For any square matrix A with real entries, prove that (1) (A + A’) is symmetric (2) (A – A’) is skew-symmetric.
Solution
Let A and B be two skew-symmetric matrices. Then,
(1) (A + A’)’ = A’ + (A’)’ [∵ (A + B)’ = A’ + B’]
= A’ = A [∵ (A’)’ = A]
= (A + A’) [∵ A + B = B + A].
Hence, (A + A’) is symmetric.
(2) (A – A’)’ = A’ – (A’)’ [∵ (A – B)’ = A’ – B’
= A’ – A [∵ (A’)’ = A]
= (A – A’).
Thus, (A – A’)’ = -(A – A’).
Hence, (A – A’) is skew-symmetric.
Multiplication of Matrices
For two given matrices A and B, we say that the product AB exists only when the number of rows in A equals the number of columns in B.
When AB exists, we say that A is conformable to B for multiplication.
Product Of Matrices
Let A = [aij]mxn and B = [bij]nxp be two matrices such that the number of columns in A equals the number of rows in B.
Then, AB exists and it is an (mxp) matrix, given by
\(A B=\left[c_{i 2}\right]_{-\times} \text {where } c_{i k}=\left(a_{i i} b_{1 k}+a_{i k} b_{2 k}+\ldots a_{i k} b_{n k}\right)=\sum_{j=1}^n a_i b_{j k}\)∴ (i,k)th element of AB
= sum of the products of corresponding elements of ith row of A and kth column of B.
Remarks For two given matrices A and B:
(1) AB may exist and BA may not exist;
(2) BA may exist and AB may not exist;
(3) AB and BA both may not exist;
(4) AB and BA both may exist.
Solved Examples
Example 1 If A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\) then show that AB and BA both exist. Find AB and BA.
Solution
Here, A is a 2 x 3 matrix and B is a 3 x 2 matrix.
∴ AB exists and it is a 2 x 2 matrix.
Let AB = \(\left[\begin{array}{ll}
c_{11} & c_{12} \\
c_{21} & c_{22}
\end{array}\right]\). Then,
c11 = (1st row of A) x (1st column of B)
= a11b11 + a12b21 + a13b31;
c12 = (1st row of A)x(2nd column of B)
= a11b12 + a12b22 + a13b32;
c21 = (2nd row of A) x (1st column of B)
= a22b11 + a22b21 + a23b31;
and c22 = (2nd row of A) x (2nd column of B)
= a21b12 + a22b22 + a23b32.
∴ AB = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a_{11} b_{11}+a_{12} b_{21}+a_{13} b_{31} & a_{11} b_{12}+a_{12} b_{22}+a_{13} b_{32} \\
a_{21} b_{11}+a_{22} b_{21}+a_{23} b_{31} & a_{21} b_{12}+a_{22} b_{22}+a_{23} b_{32}
\end{array}\right]\)
Again, B is 3 x 2 matrix and A is a 2 x 3 matrix. So, BA exists and it is a 3 x 3 matrix.
Proceeding as above, we get
BA = \(\left[\begin{array}{ll}
b_{11} & b_{12} \\
b_{21} & b_{22} \\
b_{31} & b_{32}
\end{array}\right]\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23}
\end{array}\right]\)
= \(\left[\begin{array}{lll}
b_{11} a_{11}+b_{12} a_{21} & b_{11} a_{12}+b_{12} a_{22} & b_{11} a_{13}+b_{12} a_{23} \\
b_{21} a_{11}+b_{22} a_{21} & b_{21} a_{12}+b_{22} a_{22} & b_{21} a_{13}+b_{22} a_{23} \\
b_{31} a_{11}+b_{32} a_{21} & b_{31} a_{12}+b_{32} a_{22} & b_{31} a_{13}+b_{32} a_{23}
\end{array}\right] .\)
Example 2 If A = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 4 \\
1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-1 & 3 \\
2 & 1
\end{array}\right]\), find AB. Does BA exist?
Solution
Here A is a 3 x 2 matrix and B is a 2 x 2 matrix.
Clearly, the number of columns in A equals the number of rows in B.
∴ AB exists and it is a 3 x 2 matrix.
Now, AB = \(\left[\begin{array}{rr}
2 & -1 \\
3 & 4 \\
1 & 5
\end{array}\right]\left[\begin{array}{rr}
-1 & 3 \\
2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rl}
2 \cdot(-1)+(-1) \cdot 2 & 2 \cdot 3+(-1) \cdot 1 \\
3 \cdot(-1)+4 \cdot 2 & 3 \cdot 3+4 \cdot 1 \\
1 \cdot(-1)+5 \cdot 2 & 1 \cdot 3+5 \cdot 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-4 & 5 \\
5 & 13 \\
9 & 8
\end{array}\right]\)
Further, B is a 2 x 2 matrix and A is a 3 x 2 matrix. So, the number of columns in B is not equal to the number of rows in A.
So, BA does not exist.
Example 3 Let A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\). Find AB and BA,a nd show that AB ≠ BA.
Solution
Here A is a 2 x 3 matrix and B is a 3 x 2 matrix.
So, AB exists and it is a 2 x 2 matrix.
Now, AB = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 \cdot 2+(-2) \cdot 4+3 \cdot(-2) & 1 \cdot 3+(-2) \cdot 5+3 \cdot 1 \\
(-4) \cdot 2+2 \cdot 4+5 \cdot(-2) & (-4) \cdot 3+2 \cdot 5+5 \cdot 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-12 & -4 \\
-10 & 3
\end{array}\right]\)
Again, B is a 3 x 2 matrix and A is a 2 x 3 matrix.
So, BA exists and it is a 3 x 3 matrix.
Now, BA = \(\left[\begin{array}{rr}
2 & 3 \\
4 & 5 \\
-2 & 1
\end{array}\right]\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2 \cdot 1+3 \cdot(-4) & 2 \cdot(-2)+3 \cdot 2 & 2 \cdot 3+3 \cdot 5 \\
4 \cdot 1+5 \cdot(-4) & 4 \cdot(-2)+5 \cdot 2 & 4 \cdot 3+5 \cdot 5 \\
(-2) \cdot 1+1 \cdot(-4) & (-2) \cdot(-2)+1 \cdot 2 & (-2) \cdot 3+1 \cdot 5
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-10 & 2 & 21 \\
-16 & 2 & 37 \\
-6 & 6 & -1
\end{array}\right] \text {. }\)
Hence, AB ≠ BA.
Properties of Matrix Multiplication
1. Commutativity
Matrix multiplication is not commutative in general.
Proof Let A and B be two given matrices.
If AB exists then it is quite possible that BA may not exist.
For example, if A is a 3 x 2 matrix and B is a 2 x 2 matrix then clearly, AB exists but BA does not exist.
Similarly, if BA exists then AB may not exist.
For example, if A is a 2 x 3 matrix an dB is a 2 x 2 matrix then clearly, BA exists but AB does not exist.
Further, if AB and BA both exist, then may not be comparable. For example, if A is a 2 x 3 matrix and B is a 3 x 2 matrix then clearly, AB as well ad BA exists. But, AB is a 2 x 2 matrix while BA is a 3 x 3 matrix.
Again, if AB and BA both exist and they are comparable, even then they may not be equal.
For example, if A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\) then AB and BA are both defined and each one is a 2 x 2 matrix.
But, AB = \(\left[\begin{array}{ll}
1 & 5 \\
1 & 8
\end{array}\right]\) and BA = \(\left[\begin{array}{ll}
3 & 5 \\
3 & 6
\end{array}\right]\)
This shows that AB ≠ BA.
Hence, in general, AB ≠ BA.
Remarks (1) When AB = BA, we say that A and B commute.
(2) When AB = -BA, we say that A and B anticommute.
2. Associative law
For any matrices A, B, C for which (AB)C and A(BC) both exist, we have (AB)C = A(BC).
3. Distributive laws of multiplication over addition We have:
(1) A.(B + c) = (AB + AC)
(2) (A + B).C = (AC + BC)
4. The product of two nonzero matrices can be a zero matrix.
Example Let A = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 0
\end{array}\right]\).
Then, A ≠ O and B ≠ O. But, AB = O.
Left Zero divisor If AB = O and A ≠ O then A is called a leftzerodivisior of AB.
Right Zero divisor If AB = O and B ≠ O then B is called a right zero divisor of AB.
5. If A is a given square matrix and I is an identity matrix of the same order as A then we have A.I = I.A = A.
6. If A is a given square matrix and O is the null matrix of the same order as A then O.A = A.O = O.
Positive Integral Powers of a Square Matrix
Let A be a square matrix of order n. Then, we define:
A2 = A.A;
A3 = A.A.A = A2.A;
A4 = A.A.A.A = A3.A, and so on.
∴ An = (A. A. A……n times).
Theorem 1 If A and B are square matrices of the same order then (A+B)2 = A2 + AB + BA + B2. Also, when AB = BA then (A+B)2 = A2 + 2AB + B2.
Proof Let A and B be n-rowed square matrices.
Then, clearly, (A + B) is a square matrix of order n.
So, (A + B)2 is defined.
Now, (A + B)2 = (A + B).(A + B)
= A.(A + B) + B.(A + B) [by distributive law]
= AA + AB + BA + BB [by distributive law]
=A2 + AB + BA + B2.
Hence, (A + B)2 = (A2 + AB + BA + B2).
Particular case When AB = BA
In this case, we have
(A + B)2 = (A2 + AB + AB + B2) = (A2 + 2AB + B2) [∵ BA = AB].
Theorem 2 If A and B are square matrices of the same order then (A + B)(A – B) = A2 – AB + BA – B2.
Also, when AB = BA then (A + B)(A – B) = A2 – B2.
Proof
We have
(A + B).(A – B) = A(A – B) + B(A – B) [by distributive law]
= AA – AB + BA – BB [∵ A(B – c) = AB – AC]
= A2 – AB + BA – B2.
Hence, (A + B)(A – B) = A2 – AB + BA – B2.
Particular case When AB = BA
In this case, (A + B)(A – B) = (A2 – B2) [∵ BA = AB].
Solved Examples
Example 1 If A = \(\left[\begin{array}{ll}
5 & 4 \\
2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 5 & 1 \\
6 & 8 & 4
\end{array}\right]\), find AB and BA whichever exists.
Solution
Here, A is a 2 x 2 matrix and B is a 2 x 3 matrix.
Clearly, the number of columns in A = number of rows in B.
∴ AB exists and it is a 2 x 3 matrix.
AB = \(\left[\begin{array}{ll}
5 & 4 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
3 & 5 & 1 \\
6 & 8 & 4
\end{array}\right]\)
= \(\left[\begin{array}{lll}
5 \cdot 3+4 \cdot 6 & 5 \cdot 5+4 \cdot 8 & 5 \cdot 1+4 \cdot 4 \\
2 \cdot 3+3 \cdot 6 & 2 \cdot 5+3 \cdot 8 & 2 \cdot 1+3 \cdot 4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
15+24 & 25+32 & 5+16 \\
6+18 & 10+24 & 2+12
\end{array}\right]=\left[\begin{array}{lll}
39 & 57 & 21 \\
24 & 34 & 14
\end{array}\right]\)
Again, B is a 2 x 3 matrix and A is a 2 x 2 matrix.
∴ number of columns in B ≠ number of rows in A.
So, BA does not exist.
Example 2 If A = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\) then find AB and BA. Show that AB ≠ BA.
Solution
Here, A is a 2 x 3 matrix and B is a 3 x 2 matrix.
So, number of columns in A = number of rows in B.
∴ AB exists and it is a 2 x 2 matrix.
AB = \(\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2 \cdot 2+(-1) \cdot 4+3 \cdot 1 & 2 \cdot 3+(-1) \cdot(-2)+3 \cdot 5 \\
-4 \cdot 2+5 \cdot 4+1 \cdot 1 & -4 \cdot 3+5 \cdot(-2)+1 \cdot 5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
4-4+3 & 6+2+15 \\
-8+20+1 & -12-10+5
\end{array}\right]=\left[\begin{array}{rr}
3 & 23 \\
13 & -17
\end{array}\right] .\)
Again, B is a 3 x 2 matrix and A is a 2 x 3 matrix.
So, number of columns in B = number of rows in A.
∴ BA exists and it is a 3 x 3 matrix.
BA = \(\left[\begin{array}{rr}
2 & 3 \\
4 & -2 \\
1 & 5
\end{array}\right]\left[\begin{array}{rrr}
2 & -1 & 3 \\
-4 & 5 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 \cdot 2+3 \cdot(-4) & 2 \cdot(-1)+3 \cdot 5 & 2 \cdot 3+3 \cdot 1 \\
4 \cdot 2+(-2) \cdot(-4) & 4 \cdot(-1)+(-2) \cdot 5 & 4 \cdot 3+(-2) \cdot 1 \\
1 \cdot 2+5 \cdot(-4) & 1 \cdot(-1)+5 \cdot 5 & 1 \cdot 3+5 \cdot 1
\end{array}\right]\)
= \(\left[\begin{array}{llr}
4-12 & -2+15 & 6+3 \\
8+8 & -4-10 & 12-2 \\
2-20 & -1+25 & 3+5
\end{array}\right]=\left[\begin{array}{rrr}
-8 & 13 & 9 \\
16 & -14 & 10 \\
-18 & 24 & 8
\end{array}\right]\)
Clearly, AB ≠ BA.
Example 3 If A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\) then verify that (AB)C = A(BC).
Solution
We have
AB = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
3-0-4 & 1-2+10 \\
9+0-0 & 3+4+0 \\
-6+0-2 & -2+0+5
\end{array}\right]=\left[\begin{array}{rr}
-1 & 9 \\
9 & 7 \\
-8 & 3
\end{array}\right]\)
⇒ (AB)C = \(\left[\begin{array}{rr}
-1 & 9 \\
9 & 7 \\
-8 & 3
\end{array}\right]\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-2+27 & -1+0 & 3-9 \\
18+21 & 9+0 & -27-7 \\
-16+9 & -8+0 & 24-3
\end{array}\right]=\left[\begin{array}{rrr}
25 & -1 & -6 \\
39 & 9 & -34 \\
-7 & -8 & 21
\end{array}\right] \text {. }\)
Also, BC = \(\left[\begin{array}{rr}
3 & 1 \\
0 & 2 \\
-2 & 5
\end{array}\right]\left[\begin{array}{lll}
2 & 1 & -3 \\
3 & 0 & -1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
6+3 & 3+0 & -9-1 \\
0+6 & 0+0 & 0-2 \\
-4+15 & -2+0 & 6-5
\end{array}\right]=\left[\begin{array}{rrr}
9 & 3 & -10 \\
6 & 0 & -2 \\
11 & -2 & 1
\end{array}\right]\)
⇒ (AB)C = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
3 & 2 & 0 \\
-2 & 0 & 1
\end{array}\right]\left[\begin{array}{rrr}
9 & 3 & -10 \\
6 & 0 & -2 \\
11 & -2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
9-6+22 & 3-0-4 & -10+2+2 \\
27+12+0 & 9+0-0 & -30-4+0 \\
-18+0+11 & -6+0-2 & 20-0+1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
25 & -1 & -6 \\
39 & 9 & -34 \\
-7 & -8 & 21
\end{array}\right]\)
Hence, (AB)C = A(BC).
Example 4 If A = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\) verify that A(B + C) = (AB + AC).
Solution
We have
A(B + C) = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right] \cdot\left\{\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]+\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\right\}\)
= \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
9 & -1 & -1 \\
2 & 2 & 3
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 \cdot 9+2 \cdot 2 & 3 \cdot(-1)+2 \cdot 2 & 3 \cdot(-1)+2 \cdot 3 \\
1 \cdot 9+0 \cdot 2 & 1 \cdot(-1)+0 \cdot 2 & 1 \cdot(-1)+0 \cdot 3
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
31 & 1 & 3 \\
9 & -1 & -1
\end{array}\right] \text {. }\)
Now, AB = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
1 & -2 & 5 \\
0 & 7 & 3
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 \cdot 1+2 \cdot 0 & 3 \cdot(-2)+2 \cdot 7 & 3 \cdot 5+2 \cdot 3 \\
1 \cdot 1+0 \cdot 0 & 1 \cdot(-2)+0 \cdot 7 & 1 \cdot 5+0 \cdot 3
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
3 & 8 & 21 \\
1 & -2 & 5
\end{array}\right]\)
And, AC = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 0
\end{array}\right]\left[\begin{array}{rrr}
8 & 1 & -6 \\
2 & -5 & 0
\end{array}\right]\)
= \(\left[\begin{array}{lll}
3 \cdot 8+2 \cdot 2 & 3 \cdot 1+2 \cdot(-5) & 3 \cdot(-6)+2 \cdot 0 \\
1 \cdot 8+0 \cdot 2 & 1 \cdot 1+0 \cdot(-5) & 1 \cdot(-6)+0 \cdot 0
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
28 & -7 & -18 \\
8 & 1 & -6
\end{array}\right] .\)
∴ (AB + AC) = \(\left[\begin{array}{rrr}
3 & 8 & 21 \\
1 & -2 & 5
\end{array}\right]+\left[\begin{array}{rrr}
28 & -7 & -18 \\
8 & 1 & -6
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
31 & 1 & 3 \\
9 & -1 & -1
\end{array}\right]\)
Hence, A(B + C) = (AB + AC).
Example 5 Give an example of two matrices A and B such that A ≠ O, B ≠ O and AB = BA = O.
Solution
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\). Then,
AB = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 \cdot 1+1 \cdot(-1) & 1 \cdot(-1)+1 \cdot 1 \\
1 \cdot 1+1 \cdot(-1) & 1 \cdot(-1)+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=0 .
BA = \(\left[\begin{array}{rr}
1 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 \cdot 1+1 \cdot(-1) & 1 \cdot 1+1 \cdot(-1) \\
(-1) \cdot 1+1 \cdot 1 & (-1) \cdot 1+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)=O
Hence, AB = BA = O.
Example 6 Let A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\) and I is the identity matrix of order 2. Show that (I + A) = \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution
Let \(\tan \frac{\alpha}{2}=t\)
Then, \(\cos \alpha=\frac{1-\tan ^2(\alpha / 2)}{1+\tan ^2(\alpha / 2)}=\frac{1-t^2}{1+t^2}\)
and \(\sin \alpha=\frac{2 \tan (\alpha / 2)}{1+\tan ^2(\alpha / 2)}=\frac{2 t}{1+t^2}\)
∴ \((I+A)=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+\left[\begin{array}{rr}
0 & -t \\
t & 0
\end{array}\right]=\left[\begin{array}{rr}
1 & -t \\
t & 1
\end{array}\right]\)
And, \((I-A)=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
0 & -t \\
t & 0
\end{array}\right]=\left[\begin{array}{rr}
1 & t \\
-t & 1
\end{array}\right]\)
∴ \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1 & t \\
-t & 1
\end{array}\right]\left[\begin{array}{ll}
\frac{1-t^2}{1+t^2} & \frac{-2 t}{1+t^2} \\
\frac{2 t}{1+t^2} & \frac{1-t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\frac{1-t^2}{1+t^2}+\frac{2 t^2}{1+t^2} & \frac{-2 t}{1+t^2}+\frac{t\left(1-t^2\right)}{1+t^2} \\
\frac{-t\left(1-t^2\right)}{1+t^2}+\frac{2 t}{1+t^2} & \frac{2 t^2}{1+t^2}+\frac{1-t^2}{1+t^2}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
1 & -t \\
t & 1
\end{array}\right]\)=(I+A) .
Hence, (I + A) = \((I-A) \cdot\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Example 7 If A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) then prove that \(A^n=\left[\begin{array}{rr}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right], n \in N .\)
Solution
We shall prove the result by using the principle of mathematical induction.
When n = 1, we have
\(A^1=\left[\begin{array}{rr}\cos 1 \cdot \theta & \sin 1 \cdot \theta \\
-\sin 1 \cdot \theta & \cos 1 \cdot \theta
\end{array}\right]=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right] .\)
Thus, the result is true for n = 1.
Let the result be true for n = k.
Then, \(A^k=\left[\begin{array}{rr}
\cos k \theta & \sin k \theta \\
-\sin k \theta & \cos k \theta
\end{array}\right]\)
∴ A^{k+1}
= \(A \cdot A^k=\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\left[\begin{array}{rr}
\cos k \theta & \sin k \theta \\
-\sin k \theta & \cos k \theta
\end{array}\right]\)
= \(\left[\begin{array}{rr}
\cos \theta \cos k \theta-\sin \theta \sin k \theta & \cos \theta \sin k \theta+\sin \theta \cos k \theta \\
-\sin \theta \cos k \theta-\cos \theta \sin k \theta & -\sin \theta \sin k \theta+\cos \theta \cos k \theta
\end{array}\right]\)
= \(\left[\begin{array}{rr}
\cos (\theta+k \theta) & \sin (\theta+k \theta) \\
-\sin (\theta+k \theta) & \cos (\theta+k \theta)
\end{array}\right]\)
= \(\left[\begin{array}{rr}
\cos (k+1) \theta & \sin (k+1) \theta \\
-\sin (k+1) \theta & \cos (k+1) \theta
\end{array}\right]\)
Thus, the result is true for n = (k+1), whenever it is true for n = k.
Hence, \(A^n=\left[\begin{array}{rr}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\) for all values of n ∈ N.
Example 8 Let A and B be symmetric matrices of the same order. Show that AB is symmetric if and only if AB = BA.
Solution
Since A and B are symmetric, we have A’ = A and B’ = B.
Let AB be symmetric. Then,
\((A B)^t=A B \Rightarrow B^t A^t=A B\left[(A B)^t=B^t A^t\right]\)⇒ BA = AB [∵ B’ = B and a’ = A].
Thus, AB = BA.
Conversely, let AB = BA. Then,
AB = BA ⇒ \((A B)^t=(B A)^t=A^t B^t=A B\) [∵ A’ = A and B’ = B]
⇒ AB is symmetric.
Hence, AB is symmetric ⇔ AB = BA.
Example 9 If A and B are symmetric matrices, prove that (AB – BA) is skew-symmetric.
Solution
Since A and B are symmetric, we have At = A and Bt = B.
∴ \((A B-B A)^t=(A B)^t-(B A)^t\)
= \(B^t A^t-A^t B^t\) [∵ \((A B)^t=B^t A^t \text { and }(B A)^t=A^t B^t\)]
= BA – AB [∵ \(A^t=A \text { and } B^t=B\)]
= -(AB-BA).
This shows that (AB-BA) is skew-symmetric.
Example 10 Find the matrix A such that \(\left[\begin{array}{rr}
2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right]-A=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)
Solution
Clearly, the product is a (3×3) matrix and the prefactor is a (3×2) matrix. So, A must be a (2×3) matrix.
Let A = \(\left[\begin{array}{lll}
x & y & z \\
u & v & w
\end{array}\right]\)
Then, the given equation becomes
\(\left[\begin{array}{rr}2 & -1 \\
1 & 0 \\
-3 & 4
\end{array}\right] \cdot\left[\begin{array}{lll}
x & y & z \\
u & v & w
\end{array}\right]=\left[\begin{array}{rrr}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)
⇒ \(\left[\begin{array}{ccc}
2 x-u & 2 y-v & 2 z-w \\
x & y & z \\
-3 x+4 u & -3 y+4 v & -3 z+4 w
\end{array}\right]=\left[\begin{array}{ccc}
-1 & -8 & -10 \\
1 & -2 & -5 \\
9 & 22 & 15
\end{array}\right]\)
⇒ 2x – u = -1, 2y – v = -8, 2z – w = -10, x = 1, y = -2, z = -5
⇒ x = 1, y = -2, x = -5, u = 3, v = 4 and w = 0.
Hence, A = \(\left[\begin{array}{rrr}
1 & -2 & -5 \\
3 & 4 & 0
\end{array}\right]\)
Scalar Multiplication
Let A be a given matrix and k be a nonzero real number. Then, the matrix obtained by multiplying each element of A by k is called the scalar multiple of A by k and it is denoted by kA.
Here, k is called a scalar.
If A is an (mxn) matrix then kA is also an (mxn) matrix.
If A = [aij]mxn then kA = [k.aij]mxn.
Example 1 If A = \(\left[\begin{array}{rrr}
5 & 6 & -4 \\
8 & -3 & 2
\end{array}\right]\), find (1) 3A (2) -5A (3)\frac{1}{2}A.
Solution We have:
(1) 3A = \(\left[\begin{array}{ccc}
3 \times 5 & 3 \times 6 & 3 \times(-4) \\
3 \times 8 & 3 \times(-3) & 3 \times 2
\end{array}\right]=\left[\begin{array}{rrr}
15 & 18 & -12 \\
24 & -9 & 6
\end{array}\right]\)
(2) -5A = \(\left[\begin{array}{ccc}
(-5) \times 5 & (-5) \times 6 & (-5) \times(-4) \\
(-5) \times 8 & (-5) \times(-3) & (-5) \cdot 2
\end{array}\right]=\left[\begin{array}{rrr}
-25 & -30 & 20 \\
-40 & 15 & -10
\end{array}\right] \text {. }\)
(3) \(\frac{1}{2} A=\left[\begin{array}{lll}
\frac{1}{2} \times 5 & \frac{1}{2} \times 6 & \frac{1}{2} \times(-4) \\
\frac{1}{2} \times 8 & \frac{1}{2} \times(-3) & \frac{1}{2} \times 2
\end{array}\right]=\left[\begin{array}{rrr}
\frac{5}{2} & 3 & -2 \\
4 & -\frac{3}{2} & 1
\end{array}\right] \text {. }\)
Some Properties of Scalar Multiplication
Theorem 1 If A and B are two matrices of the same order and k is a scalar then prove that k(A+B) = kA + kB.
Proof Let A = [aij]mxn and B = [bij]mxn. Then,
k(A+B) = k.([aij]mxn + [bij]mxn)
= k.[aij + bij]mxn [by definition of addition of matrices]
= [k(aij+bij)]mxn [by definition of scalar multiplication]
= [k.aij + k.bij]mxn [by distributive law]
= kA + kB.
Hence, k(A + B) = kA + kB.
Theorem 2 If A is any matrix and k1, k2 are any scalars then prove that \(\left(k_1+k_2\right) A=k_1 A+k_2 A\).
Proof
Let A = [aij]mxn. Then,
\(\left(k_1+k_2\right) A=\left(k_1+k_2\right) \cdot\left[a_{i j}\right]_{m \times n}\)= \(\left[\left(k_1+k_2\right) \cdot a_{i j}\right]_{m \times n}\) [by definition of scalar multiplication]
= \(\left[k_1 \cdot a_{i j}+k_2 \cdot a_{i j}\right]_{m \times n}\) [by distributive law]
= \(\left[k_1 \cdot a_{i j}\right]_{m \times n}+\left[k_2 \cdot a_{i j}\right]_{m \times n}\) [by definition of addition of matrices]
= \(k_1 A+k_2 A .\)
Hence, \(\left(k_1+k_2\right) A=k_1 A+k_2 A\).
Theorem 3 If A is any matrix and k1, k2 are any scalars then prove that \(k_1\left(k_2 A\right)=\left(k_1 k_2\right) A .\)
Proof
Let A = \(\left[a_{i j}\right]_{m \times n}\). Then,
\(k_1\left(k_2 A\right)=k_1\left[k_2 \cdot a_{i j}\right]_{m \times n}=\left[k_1\left(k_2 \cdot a_{i j}\right)\right]_{m \times n}\)= \(\left[\left(k_1 k_2\right) \cdot a_{i j}\right]_{m \times n}\) [by associativity of multiplication in numbers]
= \(\left(k_1 k_2\right) \cdot\left[a_{i j}\right]_{m \times n}=\left(k_1 k_2\right) A .\)
Hence, \(k_1\left(k_2 A\right)=\left(k_1 k_2\right) A .\)
Theorem 4 Let A be a given matrix and let k be a scalar. Then, prove that \((k A)^t=k A^t .\)
Proof
Let A = [aij]mxn be a given matrix and let k be a scalar. Then,
A is an (mxn) matrix ⇒ kA is an (mxn) matrix
⇒ \((k A)^t\) is an (nxm) matrix.
Also, A is an (mxn) matrix ⇒ A^t is an (nxm) matrix
⇒ \(kA^t\) is an (nxm) matrix.
Thus, \((k A)^t\) and \(\left(k A^t\right)\) are comparable matrices.
Also, (j,i)th element of \((k A)^t\) = (i,j)th element of kA
= k times (i,j)th element of \(A^t\)
= (j,i)th element of \((k A)^t\).
Thus, \((k A)^t and \left(k A^t\right)\) are comparable matrices whose corresponding elements are equal.
Hence, \((k A)^t=k A^t .\)
Theorem 5 Let A and B be two matrices such that AB exists and let c be a nonzero scalar. Then, prove that c(A x B) = (cA) x B = A x (cB).
Proof
Let A = [aij]mxn and B = [bjk]nxp. Then,
A is an (mxn) matrix, B is an (nxp) matrix
⇒ (A x B) is an (m x p) matrix
⇒ c(A x B) is an (m x p) matrix.
Again, A is an (m x n) matrix, B is an (n x p) matrix
⇒ cA is an (m x n) matrix, B is an (n x p) matrix
⇒ (cA) x B is an (m x p) matrix.
∴ c(A x B) and (cA) x B are comparable matrices.
Now, (i,k)th element of c(A x B)
= c time (i,k)th element of (A x B)
= \(c \cdot \sum_{j=1}^n a_{i j} b_{j k}\)
= \(c \cdot\left(a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+\ldots+a_{i n} b_{n k}\right)\)
= \(\left(c a_{i 1}\right) b_{1 k}+\left(c a_{i 2}\right) b_{2 k}+\ldots+\left(c a_{i n}\right) b_{n k}\)
= (i,j)th element of (cA) x B.
Thus, c(A x B) are (cA) x B are comparable and their corresponding elements are equal.
Hence, c(A x B) = (cA x B).
Again, A is an (m x n) matrix and B is an (n x p) matrix
⇒ A is an (m x n) matrix and cB is an (n x p) matrix
⇒ A x (cB) is an (m x p) matrix.
Thus, c(A x B) and A x (cB) are comparable.
Also, (i,k)th element of A x (cB)
= \(\sum_{j=1}^n a_{i j}\left(c b_{j k}\right)\)
= \(a_{i 1}\left(c b_{1 k}\right)+a_{i 2}\left(c b_{2 k}\right)+\ldots+a_{i n}\left(c b_{n k}\right)\)
= \(c\left(a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+\ldots+a_{i n} b_{n k}\right)\)
= \(c \sum_{j=1}^n a_{i j} b_{j k}\) = (i,k)th element of c(A x B).
∴ c(A x B) = A x (cB)
Hence, c(A x B) = (cA) x B = A x (cB).
Summary of properties of Matrix Operations
1. (1) \(\text { (i) }\left(A^t\right)^t=A\)
(2) \((A+B)^t=\left(A^t+B^t\right) .\)
(3) \((A \times B)^t \neq\left(A^t \times B^t\right) \text { and }(A \times B)^t \neq\left(B^t \times A^t\right) \text {. }\)
2. (1) (A + B) = (B + A)
(2) (A + B) + C = A + (B + C)
(3) For all comparable matrices A there exists a null matrix O such that A + O = O + A = A.
3. (1) (A x B) ≠ (B x A).
(2) (A x B) x C = A x (B x C)
(3) A x (B + C) = (A x B) + (A x C)
(4) (A + B) x C = (A x C) + (B x C)
(5) For all square matrices A of the same order there exists a unit matrix I such that (A x I) = (I x A) = A.
4. (1) (C A)^t=C A^{\prime} \text {. }
(2) c(A + B) = (cA + cB)
(3) c(A x B) = (cA x B) = A x (cB).
(4) ∃ a matrix S such that St = S(called a symmetric matrix)
More Solved Examples
Example 2 If A = \(\left[\begin{array}{rr}
3 & 5 \\
7 & -9
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
6 & -4 \\
2 & 3
\end{array}\right]\), find (4A – 3B).
Solution
We have: 4A – 3B = 4A + (-3)B.
Now, \(4 A=\left[\begin{array}{ll}
4 \times 3 & 4 \times 5 \\
4 \times 7 & 4 \times(-9)
\end{array}\right]=\left[\begin{array}{rr}
12 & 20 \\
28 & -36
\end{array}\right]\)
and (-3)B = \(\left[\begin{array}{ll}
(-3) \times 6 & (-3) \times(-4) \\
(-3) \times 2 & (-3) \times 3
\end{array}\right]=\left[\begin{array}{rr}
-18 & 12 \\
-6 & -9
\end{array}\right] .\)
∴ 4A – 3B = 4A + (-3)B-B
= \(\left[\begin{array}{rr}
12 & 20 \\
28 & -36
\end{array}\right]+\left[\begin{array}{rr}
-18 & 12 \\
-6 & -9
\end{array}\right]\)
= \(\left[\begin{array}{cc}
12+(-18) & 20+12 \\
28+(-6) & (-36)+(-9)
\end{array}\right]=\left[\begin{array}{rr}
-6 & 32 \\
22 & -45
\end{array}\right] \text {. }\)
Hence, (4A – 3B) = \(\left[\begin{array}{rr}
-6 & 32 \\
22 & -45
\end{array}\right]\)
Example 3 Let A = diag[3,-5,7] and B = diag[-1, 2,4]. Find (1) (A + B), (2) (A – B), (3) -5A, (4) (2A + 3B).
Solution
We have
\(A=\left[\begin{array}{rrr}3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right] \text { and } B=\left[\begin{array}{rrr}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]\)
∴ (1) \(A+B=\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]+\left[\begin{array}{rrr}
-1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{rrr}
2 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & 11
\end{array}\right]\)
(2) (A – B) = A + (-B)
= \(\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]+\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -4
\end{array}\right]=\left[\begin{array}{rrr}
4 & 0 & 0 \\
0 & -7 & 0 \\
0 & 0 & 3
\end{array}\right]\)
(3) -5A = (-5).A = \((-5) \cdot\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & 7
\end{array}\right]=\left[\begin{array}{rrr}
-15 & 0 & 0 \\
0 & 25 & 0 \\
0 & 0 & -35
\end{array}\right]\)
(4) 2A + 3B = \(\left[\begin{array}{rrr}
6 & 0 & 0 \\
0 & -10 & 0 \\
0 & 0 & 14
\end{array}\right]+\left[\begin{array}{rrr}
-3 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 12
\end{array}\right]=\left[\begin{array}{rrr}
3 & 0 & 0 \\
0 & -4 & 0 \\
0 & 0 & 26
\end{array}\right]\)
Example 4 Simplify \(\cos \theta \cdot\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta \cdot\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \text {. }\)
Solution
We have
\(\cos \theta \cdot\left[\begin{array}{rr}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta \cdot\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \text {. }\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta & \sin \theta \cos \theta \\
-\sin \theta \cos \theta & \cos ^2 \theta
\end{array}\right]+\left[\begin{array}{cc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta+\sin ^2 \theta & \sin \theta \cos \theta+(-\sin \theta \cos \theta) \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] .\)
Example 5 If \(2\left[\begin{array}{cc}
x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rc}
7 & 14 \\
15 & 14
\end{array}\right]\), find the values of x and y.
Solution
We have
\(2\left[\begin{array}{cc}x & 5 \\
7 & y-3
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
7 & 14 \\
15 & 14
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 x & 10 \\
14 & 2 y-6
\end{array}\right]+\left[\begin{array}{ll}
3 & 4 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rc}
7 & 14 \\
15 & 14
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 x+3 & 14 \\
15 & 2 y-4
\end{array}\right]=\left[\begin{array}{rr}
7 & 14 \\
15 & 14
\end{array}\right]\)
⇒ 2x + 3 = 7 and 2y – 4 = 14
⇒ x = 2 and y = 9.
The values of x and y are 3 and 9
Example 6 Find matrix X such that \(X+\left[\begin{array}{rr}
4 & 6 \\
-3 & 7
\end{array}\right]=\left[\begin{array}{ll}
3 & -6 \\
5 & -8
\end{array}\right]\)
Solution
Let A = \(\left[\begin{array}{rr}
4 & 6 \\
-3 & 7
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -6 \\
5 & -8
\end{array}\right]\)
Then, the given matrix equation is X + A = B.
Now, X + A = B ⇒ X = B – A = B + (-A)
= \(\left[\begin{array}{rr}
3 & -6 \\
5 & -8
\end{array}\right]+\left[\begin{array}{rr}
-4 & -6 \\
3 & -7
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3+(-4) & -6+(-6) \\
5+3 & -8+(-7)
\end{array}\right]=\left[\begin{array}{rr}
-1 & -12 \\
8 & -15
\end{array}\right] \text {. }\)
Hence, X = \(\left[\begin{array}{rr}
-1 & -12 \\
8 & -15
\end{array}\right]\)
Example 7 Find a matrix X such that 2A + B + X = O, where A = \(\left[\begin{array}{rr}
-1 & 2 \\
3 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right] \text {. }\)
Solution
We have
2A + B + X = O ⇒ X = -(2A + B).
Now, (2A + B) = \(2 \cdot\left[\begin{array}{rr}
-1 & 2 \\
3 & 4
\end{array}\right]+\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rr}
-2 & 4 \\
6 & 8
\end{array}\right]+\left[\begin{array}{rr}
3 & -2 \\
1 & 5
\end{array}\right]\)
= \(\left[\begin{array}{rl}
-2+3 & 4+(-2) \\
6+1 & 8+5
\end{array}\right]=\left[\begin{array}{rr}
1 & 2 \\
7 & 13
\end{array}\right]\)
∴ X = -(2A + B) = \(\left[\begin{array}{cc}
-1 & -2 \\
-7 & -13
\end{array}\right]\)
Example 8 Find matrices X and Y, if X + Y = \(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]\) and X – Y = \(\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\).
Solution
Adding the given matrices, we get
(X + Y) + (X – Y) = \(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]+\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\)
⇒ \(2 X=\left[\begin{array}{ll}
5+3 & 2+6 \\
0+0 & 9+(-1)
\end{array}\right]\)
⇒ \(2 X=\left[\begin{array}{ll}
8 & 8 \\
0 & 8
\end{array}\right]\) ⇒ \(X=\frac{1}{2} \cdot\left[\begin{array}{ll}
8 & 8 \\
0 & 8
\end{array}\right]=\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right]\)
On substracting the given matrices, we get
(X + Y) – (X – Y)=\(\left[\begin{array}{ll}
5 & 2 \\
0 & 9
\end{array}\right]-\left[\begin{array}{rr}
3 & 6 \\
0 & -1
\end{array}\right]\)
⇒ \(2 Y=\left[\begin{array}{ll}
5-3 & 2-6 \\
0-0 & 9-(-1)
\end{array}\right]=\left[\begin{array}{ll}
2 & -4 \\
0 & 10
\end{array}\right]\)
⇒ \(Y=\frac{1}{2}\left[\begin{array}{rr}
2 & -4 \\
0 & 10
\end{array}\right]=\left[\begin{array}{rr}
1 & -2 \\
0 & 5
\end{array}\right] .\)
Hence, X = \(\left[\begin{array}{ll}
4 & 4 \\
0 & 4
\end{array}\right] and Y = \left[\begin{array}{rr}
1 & -2 \\
0 & 5
\end{array}\right] \text {. }\)
Matrix Polynomial Let f(x) = \(a_0 x^m+a_1 x^{m-1}+a_2 x^{m-2}+\ldots+a_{m-1} x+a_m\) be a polynomial of degree m and let A be a square matrix of order n. Then, the corresponding matrix polynomial is:
f(A) = \(a_0 A^{m+}+a_1 A^{m-1}+a_2 A^{m-2}+\ldots+a_{m-1} A+a_m I\), where I is a unit matrix of order n.
Example 9 If f(x) = \(x^2-5 x+7\) and A = \(\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\), find f(A).
Solution
f(x) = \(x^2-5 x+7\) ⇒ f(A) = \(A^2-5 A+7 I\)
Now, \(A^2=\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\)
= \(\left[\begin{array}{rr}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]\)
-5A = \(\left[\begin{array}{ll}
(-5) \cdot 3 & (-5) \cdot 1 \\
(-5) \cdot(-1) & (-5) \cdot 2
\end{array}\right]=\left[\begin{array}{rr}
-15 & -5 \\
5 & -10
\end{array}\right]\)
7I = \(7 \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] .\)
∴ f(A) = \(A^2-5 A+7 I\)
= \(\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]+\left[\begin{array}{rr}
-15 & -5 \\
5 & -10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right]\)
= \(\left[\begin{array}{cc}
8+(-15)+7 & 5+(-5)+0 \\
-5+5+0 & 3+(-10)+7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
Hence, f(A) = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0 .\)
Example 10 If A = \(\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that \(A^2-5 A-14 I=0 .\)
Solution
We have
\(A^2=\left[\begin{array}{rr}3 & -5 \\
-4 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3 \cdot 3+(-5)(-4) & 3 \cdot(-5)+(-5) \cdot 2 \\
(-4) \cdot 3+2 \cdot(-4) & (-4) \cdot(-5)+2 \cdot 2
\end{array}\right]=\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]\)
-5A = \((-5)\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right] \text {; }\)
-14I = \((-14)\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right] \text {. }\)
∴ \(A^2-5 A-14 I=A^2+(-5) A+(-14 I)\)
= \(\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]+\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right]+\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right]\)
= \(\left[\begin{array}{cc}
29+(-15)+(-14) & -25+25+0 \\
-20+20+0 & 24+(-10)+(-14)
\end{array}\right]\)
= \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0 \text {. }\)
Hence, \(A^2-5 A-14 I=0 .\)
Example 11 If A = \(\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that \(A^2=8 A+k I\).
Solution
We have
\(A^2=\left[\begin{array}{rr}1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]=\left[\begin{array}{rl}
1-0 & 0+0 \\
-1-7 & 0+49
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]\)
(8A + kI) = \(8 \cdot\left[\begin{array}{rr}
1 & 0 \\
-1 & 7
\end{array}\right]+k \cdot\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rr}
8 & 0 \\
-8 & 56
\end{array}\right]+\left[\begin{array}{ll}
k & 0 \\
0 & k
\end{array}\right]=\left[\begin{array}{cc}
8+k & 0 \\
-8 & 56+k
\end{array}\right]\)
∴ \(A^2=8 A+k I\) ⇒ \(\left[\begin{array}{rr}
1 & 0 \\
-8 & 49
\end{array}\right]=\left[\begin{array}{cc}
8+k & 0 \\
-8 & 56+k
\end{array}\right]\)
⇒ 8 + k = 1 and 56 + k = 49
⇒ k = -7.
Hence, k = -7.
Example 12 If A = \(\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]\), prove that for all n ∈ N, \((a I+b A)^n=a^n I+n d^{n-1} b A\), where I is the identity matrix of order 2.
Solution
We shall prove the result by mathematical induction.
When n = 1, we have:
LHS = \((a I+b A)^1=(a I+b A)=\left(a^1 I+1 a^0 b A\right)\) = RHS.
So, the result is true for n = 1.
Let it be true for n = m, so that
\((a I+b A)^m=a^m I+m a^{m-1} b A\) …(1)
∴ \((a I+b A)^{m+1}\)
= \((a I+b A) \cdot(a I+b A)^m=(a I+b A) \cdot\left(a^m I+m a^{m-1} b A\right)\) [using(1)]
= \(a I\left(a^m I+m a^{m-1} b A\right)+b A\left(a^m I+m a^{m-1} b A\right)\)
= \(a^{m+1} I+m d^m b A+a^m b A+m a^{m-1} b^2 A^2\) [∵ II = I, IA = A = AI]
= \(a^{m+1} I+(m+1) a^m b A\) [∵ \(A^2=\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right] \times\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=O]\).
Thus, whenever the result is true for n = m, then it is also true for n = (m + 1).
Hence, by mathematical induction, it is true for all n ∈ N.
Theorem 6 If A is a symmetric matrix, then prove that kA is symmetric.
Proof Since A is symmetric, we have \(A^t=A .\)
∴ \((k A)^t=k A^t=k A\) [∵ \(A^t=A\)].
Hence, (kA) is symmetric.
Theorem 7 If A is a skew-symmetric matrix, then prove that kA is skew-symmetric.
Proof Since A is skew-symmetric, we have \(A^t=-A .\)
∴ \((k A)^t=k \cdot A^t=k \cdot(-A)=-(k A)\) [∵ \(A^t=-A\)].
Hence, (kA) is skew-symmetric.
Theorem 8 Prove that every square matrix is expressible as the sum of a symmetric matrix and a skew-symmetric matrix.
Proof Let A be any square matrix. Then, we can write
\(A=\frac{1}{2}\left(A+A^t\right)+\frac{1}{2}\left(A-A^t\right)=P+Q \text { (say). }\)Then, it is easy to verify that P is symmetric and Q is skew-symmetric.
Hence, the theorem follows.
Example 13 Express the matrix A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\) as the sum of a symmetric matrix and a skew-symmetric matrix.
Solution
We know that \(A=\frac{1}{2}\left(A+A^{\dagger}\right)+\frac{1}{2}\left(A-A^{\dagger}\right)=P+Q\), where P is a symmetric and Q is skew-symmetric.
∴ \(P=\frac{1}{2}\left(A+A^t\right)=\frac{1}{2} \cdot\left\{\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]+\left[\begin{array}{rr}
3 & 1 \\
-4 & -1
\end{array}\right]\right\}\)
= \(\frac{1}{2} \cdot\left[\begin{array}{cc}
3+3 & -4+1 \\
1+(-4) & -1+(-1)
\end{array}\right]=\frac{1}{2} \cdot\left[\begin{array}{rr}
6 & -3 \\
-3 & -2
\end{array}\right]=\left[\begin{array}{cc}
3 & \frac{-3}{2} \\
\frac{-3}{2} & -1
\end{array}\right]\)
And, \(Q=\frac{1}{2}\left(A-A^t\right)=\frac{1}{2} \cdot\left\{\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]-\left[\begin{array}{rr}
3 & 1 \\
-4 & -1
\end{array}\right]\right\}\)
= \(\frac{1}{2} \cdot\left[\begin{array}{cc}
3-3 & -4-1 \\
1-(-4) & -1-(-1)
\end{array}\right]=\frac{1}{2} \cdot\left[\begin{array}{cc}
0 & -5 \\
1+4 & -1+1
\end{array}\right]\)
= \(\frac{1}{2} \cdot\left[\begin{array}{rr}
0 & -5 \\
5 & 0
\end{array}\right]=\left[\begin{array}{rr}
0 & -\frac{5}{2} \\
\frac{5}{2} & 0
\end{array}\right] \text {. }\)
Hence, A = P + Q, where P is symmetric and Q is skew-symmetric.
Example 14 Express the matrix A = \(\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]\) as the sum of a symmetric matrix and a skew-symmetric matrix.
Solution
We know that \(A=\frac{1}{2}\left(A+A^t\right)+\frac{1}{2}\left(A-A^t\right)=P+Q\), where P is symmetric and Q is skew-symmetric.
Now, \(P=\frac{1}{2}\left(A+A^t\right)\)
= \(\frac{1}{2} \cdot\left\{\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]+\left[\begin{array}{rrr}
1 & -6 & -4 \\
3 & 8 & 6 \\
5 & 3 & 5
\end{array}\right]\right\}\)
= \(\frac{1}{2} \cdot\left[\begin{array}{rrr}
2 & -3 & 1 \\
-3 & 16 & 9 \\
1 & 9 & 10
\end{array}\right]=\left[\begin{array}{rrr}
1 & -\frac{3}{2} & \frac{1}{2} \\
-\frac{3}{2} & 8 & \frac{9}{2} \\
\frac{1}{2} & \frac{9}{2} & 5
\end{array}\right]\)
And, \(Q=\frac{1}{2}\left(A-A^t\right)\)
= \(\frac{1}{2} \cdot\left\{\left[\begin{array}{rrr}
1 & 3 & 5 \\
-6 & 8 & 3 \\
-4 & 6 & 5
\end{array}\right]-\left[\begin{array}{rrr}
1 & -6 & -4 \\
3 & 8 & 6 \\
5 & 3 & 5
\end{array}\right]\right\}\)
= \(\frac{1}{2} \cdot\left[\begin{array}{rrr}
0 & 9 & 9 \\
-9 & 0 & -3 \\
-9 & 3 & 0
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{9}{2} & \frac{9}{2} \\
-\frac{9}{2} & 0 & -\frac{3}{2} \\
-\frac{9}{2} & \frac{3}{2} & 0
\end{array}\right]\)
Hence, A = P + Q, where P is symmetric and Q is skew-symmetric.
Elementary Operations on Matrices
Given below are three row operations and three column operations on a matrix, which are called elementary operations or transformations.
Equivalent Matrices Two matrices A and B are said to be equivalent if one is obtained from the other by one or more elementary operations and we write, A ~ B.
Three Elementary Row Operations
(1) Interchange of any two rows The interchange of ith and jth rows is denoted by Ri ⟷ Rj.
Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{2} & 4 & 6 \\
5 & -3 & 7
\end{array}\right]\)
Applying R2 ⟷ R3, we get A ~ \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
5 & -3 & 7 \\
\sqrt{2} & 4 & 6
\end{array}\right] \text {. }\)
(2) Multiplication of the elements of a row by a nonzero number Suppose each element of ith row of a given matrix is multiplied by a nonzero number k.
Then, we denote it by Ri → kRi.
Example Let A = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
\sqrt{3} & -5 & 6 \\
1 & 8 & 4
\end{array}\right]\)
Applying R2 → 4R2, we get A ~ \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
4 \sqrt{3} & -20 & 24 \\
1 & 8 & 4
\end{array}\right]\)
(3) Multiplying each element of a row by a nonzero number and then adding them to the corresponding elements of another row Suppose each element of jth row of a matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith row.
We dentoe it by Ri → Ri + kRj.
Example Let A = \(\left[\begin{array}{rrr}
2 & -1 & 5 \\
-3 & 4 & \sqrt{2} \\
7 & 6 & 3
\end{array}\right]\)
Applying R1 → R1 + 2R3, we get A ~ \(\left[\begin{array}{rrr}
16 & 11 & 11 \\
-3 & 4 & \sqrt{2} \\
7 & 6 & 3
\end{array}\right]\)
Three Elementary Column Operations
(1) Interchange of any two columns The interchange of ith and jth columns is denoted by \(c_i \leftrightarrow c_j .\)
Example Let A = \(\left[\begin{array}{rrr}
2 & 1 & -3 \\
-1 & 5 & 4 \\
6 & 3 & \frac{1}{2}
\end{array}\right]\)
Applying \(c_1 \leftrightarrow c_2\), we get A ~ \(\left[\begin{array}{rrr}
1 & 2 & -3 \\
5 & -1 & 4 \\
3 & 6 & \frac{1}{2}
\end{array}\right] \text {. }\)
(2) Multiplying each element of a column by a nonzero number Suppose each element of ith column of matrix A is multiplied by a nonzero number k.
Then, we write, \(\mathcal{C}_i \rightarrow k \mathcal{C}_i\)
Example Let A = \(\left[\begin{array}{rrr}
3 & 1 & -5 \\
\sqrt{2} & -2 & 4 \\
6 & 2 & 8
\end{array}\right]\)
Applying \(\mathcal{C}_3 \rightarrow 2 \mathcal{C}_3\), we get A ~ \(\left[\begin{array}{rrr}
3 & 1 & -10 \\
\sqrt{2} & -2 & 8 \\
6 & 2 & 16
\end{array}\right]\)
(3) Multiplying each element of a column of a given matrix A by a nonzero number and then adding to the corresponding elements of another column Suppose each element of ith column of a given matrix A is multiplied by a nonzero number k and then added to the corresponding elements of ith column.
Then we write, \(C_i \rightarrow C_i+k C_j .\)
Example Let A = \(\left[\begin{array}{rrr}
2 & 0 & 4 \\
-1 & 3 & 1 \\
5 & -2 & 6
\end{array}\right]\)
Applying \(C_3 \rightarrow C_3+2 C_1\), we get A ~ \(\left[\begin{array}{rrr}
2 & 0 & 8 \\
-1 & 3 & -1 \\
5 & -2 & 16
\end{array}\right]\)
Invertible Matrices A square matrix A of order n is said to be invertible if there exists a square matrix B of order n such that AB = BA = I.
Also, then B is called the inverse of A and we write, \(A^{-1}=B\).
Example Let A = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]\). Then,
AB = \(\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]=\left[\begin{array}{rr}
6-5 & -15+15 \\
2-2 & -5+6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\),
BA = \(\left[\begin{array}{rr}
2 & -5 \\
-1 & 3
\end{array}\right]\left[\begin{array}{ll}
3 & 5 \\
1 & 2
\end{array}\right]=\left[\begin{array}{rr}
6-5 & 10-10 \\
-3+3 & -5+6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=I\)
∴ AB = BA = I.
Hence, \(A^{-1}=B\).
Theorem 1 (Uniqueness of Inverse) Every invertible square matrix has a unique inverse.
Proof
Let A be an invertible square matrix of order n.
If possible, let B as well as C be the inverse of A.
Then, AB = BA = I and AC = CA = I.
Now, AC = I ⇒ B(AC) = B.I = B,
BA = I ⇒ (BA)C = I.C = C.
But, B(AC) = (BA)C [by associative law of multiplication]
∴ B = C.
Hence, an invertible matrix has a unique inverse.
Inverse Of A Matrix By Elementary Row Operations
Let A be a square matrix of order n.
We can write, A = I.A …(1)
Now, let a sequence of elementary row operations reduce A on LHS of (1) to I and I on RHS of (1) to a matrix B.
Then, I = BA ⇒ \(I \cdot A^{-1}=(B A) A^{-1}=B\left(A A^{-1}\right)=B I\)
⇒ \(A^{-1}=B\).
We can summarise the above method as given below.
Method Step 1. Write A = I.A.
Step 2. By using elementary row operations on A, transform it into a unit matrix.
Step 3. In the same order we apply elementary operations on I to convert it into a matrix B.
Step 4. Then, \(A^{-1}=B\).
Remark If on applying one or more elementary row operations on A, we obtain all zeros in one or more rows, then we say that \(A^{-1}\) does not exist.
Solved Examples
Example 1 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{ll}
1 & -2 \\
2 & -6
\end{array}\right]\)
Solution
We have
\(\left[\begin{array}{cc}1 & -2 \\
2 & -6
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{ll}
1 & -2 \\
0 & -2
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-2 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rr}
1 & -2 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
1 & \frac{-1}{2}
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
3 & -1 \\
1 & -\frac{1}{2}
\end{array}\right] \cdot A\)
Hence, \(A^{-1}=\left[\begin{array}{rr}
3 & -1 \\
1 & -\frac{1}{2}
\end{array}\right]\)
Example 2 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]\)
Solution
We have
A = \(\left[\begin{array}{rr}
3 & -1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{cc}
-1 & 1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rr}
1 & -1 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
-1 & -1 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{ll}
1 & -1 \\
0 & -2
\end{array}\right]=\left[\begin{array}{cc}
-1 & -1 \\
-4 & -3
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-1 & -1 \\
2 & \frac{3}{2}
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & \frac{3}{2}
\end{array}\right] \cdot A\)
Hence, \(A^{-1}=\left[\begin{array}{ll}
1 & \frac{1}{2} \\
2 & \frac{3}{2}
\end{array}\right] \text {. }\)
Example 3 If A = \(\left[\begin{array}{rr}
6 & -3 \\
-2 & 1
\end{array}\right]\), show that \(A^{-1}\) does not exist.
Solution
We have
\(\left[\begin{array}{rr}6 & -3 \\
-2 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
-2 & 1
\end{array}\right]=\left[\begin{array}{ll}
\frac{1}{6} & 0 \\
0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
0 & 0
\end{array}\right]=\left[\begin{array}{rr}
\frac{1}{6} & 0 \\
\frac{1}{3} & 1
\end{array}\right] \cdot A\)
Thus, we have all zeros in second row of the left-hand side matrix. Hence, \(A^{-1}\) does not exist.
Example 4 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution
We have
\(\left[\begin{array}{rrr}1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
0 & 9 & -11 \\
0 & -1 & 4
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
3 & 1 & 0 \\
-2 & 0 & 1
\end{array}\right] \cdot A\)
R_2 \rightarrow R_2+3 R_1 \\
R_3 \rightarrow R_3-2 R_1
\end{array}\right]\)
⇒ \(\left[\begin{array}{rrr}
1 & 3 & -2 \\
0 & -1 & 4 \\
0 & 9 & -11
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 0 & 1 \\
3 & 1 & 0
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & -1 & 4 \\
0 & 0 & 25
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 0 & 3 \\
-2 & 0 & 1 \\
-15 & 1 & 9
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1+3 R_2 \\
R_3 \rightarrow R_3+9 R_2
\end{array}\right]\)
⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & 1 & -4 \\
0 & 0 & 25
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 0 & 3 \\
2 & 0 & -1 \\
-15 & 1 & 9
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 0 & 10 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
-5 & 0 & 3 \\
2 & 0 & -1 \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right]\)
⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & \frac{-2}{5} & \frac{-3}{5} \\
\frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1-10 R_3 \\
R_2 \rightarrow R_2+4 R_3
\end{array}\right]\)
Hence, \(A^{-1}=\left[\begin{array}{ccc}
1 & \frac{-2}{5} & \frac{-3}{5} \\
\frac{-2}{5} & \frac{4}{25} & \frac{11}{25} \\
\frac{-3}{5} & \frac{1}{25} & \frac{9}{25}
\end{array}\right] \text {. }\)
Example 5 By using elementary row operations, find the inverse of the matrix A = \(\left[\begin{array}{rrr}
3 & -1 & -2 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right] .\)
Solution
We have
\(\left[\begin{array}{rrr}3 & -1 & -2 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
2 & 0 & -1 \\
3 & -5 & 0
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 2 & 1 \\
0 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
-2 & 3 & 0 \\
-3 & 3 & 1
\end{array}\right] \cdot A\)
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3-3 R_1
\end{array}\right]\)
⇒ \(\left[\begin{array}{rrr}
1 & -1 & -1 \\
0 & 1 & \frac{1}{2} \\
0 & -2 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & -1 & 0 \\
-1 & \frac{3}{2} & 0 \\
-3 & 3 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{2} \\
0 & 1 & \frac{1}{2} \\
0 & 0 & 4
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{1}{2} & 0 \\
-1 & \frac{3}{2} & 0 \\
-5 & 6 & 1
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1+R_2 \\
R_3 \rightarrow R_3+2 R_2
\end{array}\right\}\)
⇒ \(\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{2} \\
0 & 1 & \frac{1}{2} \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
0 & \frac{1}{2} & 0 \\
-1 & \frac{3}{2} & 0 \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
\frac{-5}{8} & \frac{5}{4} & \frac{1}{8} \\
\frac{-3}{8} & \frac{3}{4} & \frac{-1}{8} \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1+\frac{1}{2} R_3 \\
R_2 \rightarrow R_2-\frac{1}{2} R_3
\end{array}\right\} .\)
Hence, \(A^{-1}=\left[\begin{array}{ccc}
\frac{-5}{8} & \frac{5}{4} & \frac{1}{8} \\
\frac{-3}{8} & \frac{3}{4} & \frac{-1}{8} \\
\frac{-5}{4} & \frac{3}{2} & \frac{1}{4}
\end{array}\right] \text {. }\)
Example 6 By using elementary row transformations, find the invers of the matrix A = \(\left[\begin{array}{rrr}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution
We have
\(\left[\begin{array}{rrr}2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
2 & 0 & -1 \\
1 & 1 & 2 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
2 & 0 & -1 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
0 & -2 & -5 \\
0 & 1 & 3
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
5 & -2 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 1 & 2 \\
0 & 1 & 3 \\
0 & -2 & -5
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & 0 \\
0 & 0 & 1 \\
5 & -2 & 0
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 3 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 1 & -1 \\
0 & 0 & 1 \\
5 & -2 & 2
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1-R_2 \\
R_3 \rightarrow R_3+2 R_2
\end{array}\right]\)
⇒ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right] \cdot A\)
R_1 \rightarrow R_1+R_3 \\
R_2 \rightarrow R_2-3 R_3
\end{array}\right]\)
Hence, \(A^{-1}=\left[\begin{array}{rrr}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
Example 7 If A = \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & -1 \\
-1 & -2 & 2
\end{array}\right]\), show that \(A^{-1}\) does not exist.
Solution
We have
\(\left[\begin{array}{rrr}1 & -1 & 1 \\
2 & 1 & -1 \\
-1 & -2 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \cdot A\)
⇒ \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
0 & 3 & -3 \\
0 & -3 & 3
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
1 & 0 & 1
\end{array}\right], A\)
R_2 \rightarrow R_2-2 R_1 \\
R_3 \rightarrow R_3+R_1
\end{array}\right]\)
⇒ \(\left[\begin{array}{rrr}
1 & -1 & 1 \\
0 & 3 & -3 \\
0 & 0 & 0
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
-1 & 1 & 1
\end{array}\right] \cdot A\)
Thus, we have all zeros in 3rd row of the left-hand side matrix.
Hence, A^{-1} does not exist.