EQUALITY OF MATRICES
DEFINITION: Two matrices A = [aij] mxn And B = [bij] mxn are said to be equal if
1. m = r, i.e., the number of rows in A equals the number of rows in B
2. n = s, i.e., the number of columns in A equals the number of columns in B
3. aI j = bij for i = 1, 2, …, in and j = 1, 2, …., n
If two matrices A and B are equal, we write A = B, otherwise we write A # B.
ILLUSTRATION 1: The matrices \(A=\left[ \begin{matrix}3 & 2 & 1 \\x & y & 5\\1 & -1 & 4 \\\end{matrix} \right]\) and \(B=\left[ \begin{matrix}3 & 2 & 1\\-1 & 0 & 5 \\-1 & -1 & Z \\\end{matrix} \right]\) are equal if x = -1, y = 0 and z = 4.
ILLUSTRATION 2: If \(\left[ \begin{matrix}x-y & 2x+z \\2x-y &3z+w \\\end{matrix} \right]=\left[ \begin{matrix}-1 & 5 \\0 & 13 \\\end{matrix} \right]\) find x, y, z, w.
SOLUTION: Since the corresponding elements of two equal matrices are equal. Therefore,
\(\left[ \begin{matrix}x-y & 2x+z \\2x-y & 3z+w \\\end{matrix} \right]=\left[ \begin{matrix}-1 & 5 \\0 & 13 \\\end{matrix} \right]\).
x – y = -1, 2x – V z = 5, 2x – y = 0, 3z w = 13.
Solving the equation x – y = -1 and 2x – y = 0 as simultaneous linear equations, we get x = 1, y = 2.
Now, putting x = 1 in 2x + z = 5, we get z = 3.
Substituting z = 3 in 3z + w = 13, we obtain w = 4.
Thus, the given matrices are equal if x = 1, y = 2, z = 3 and w = 4.
ILLUSTRATION 3: Find the values of x, y, z and a which satisfy the matrix equation.
\(\left[ \begin{matrix}x+3 & 2y+x \\z-1 & 4a-6 \\\end{matrix}\right]=\left[ \begin{matrix}0 & -7 \\3 & 2a \\\end{matrix} \right]\).
PROOF: Since the corresponding elements of two equal matrices are equal.
Therefore, x + 3 = 0, 2y x = – 7, z – 1 = 3 and 4a – 6 = 2a.
Solving these equations, we get a = 3, x = – 3, y = – 2, z = 4.
ADDITION OF MATRICES
DEFINITION: Let A, B be two matrices, each of order m x n. Then their sum A+B is a matrix of order m x n and is obtained by adding the corresponding element of A and B.
Thus, if A = [aij] mxn and B = [bij] mxn are two matrices of the same order, their sum A+B is defined to be the matrix of order m x n such that (A + B) ij = + bij for i = 1, 2, …, m and j = 1, 2, …, n.
NOTE: The sum of two matrices is defined only when they are of the same order.
ILLUSTRATION 1:
\(A=\left[ \begin{align}& \begin{matrix}1 & 2 & 3 \\\end{matrix} \\& \begin{matrix}4 & 5 & 6\\\end{matrix}\\\end{align}\right]\),
\(B=\left[ \begin{align}& \begin{matrix}6 & 5 & 4 \\\end{matrix}\\ & \begin{matrix}3 & 2 & 1\\\end{matrix}\\\end{align}\right]\),
then \(A+B=\left[ \begin{align}& \begin{matrix}1+6 & 2+5 & 3+4 \\\end{matrix} \\ & \begin{matrix}4+3 & 5+2 & 6+1 \\\end{matrix} \\\end{align} \right]\)=\(\left[ \begin{align}& \begin{matrix}7 & 7 & 7 \\\end{matrix} \\& \begin{matrix}7 & 7 & 7 \\\end{matrix} \\\end{align} \right]\).
ILLUSTRATION 2: If \(A=\left[ \begin{align}& \begin{matrix}1 & 2 & 3\\\end{matrix} \\& \begin{matrix}4 & 5 & 6 \\\end{matrix} \\\end{align} \right]\), \(B=\left[ \begin{matrix}-1 & 2 & 1 \\3 & 2 & 1 \\2 & 5 & -5 \\\end{matrix} \right]\), then A + B is not defined, because A and B are not of the same order.
ILLUSTRATION 3: For the following pairs of matrices A-FB is not defined because they are of different orders:
1. \(A=\left[ \begin{matrix}1 & -1 \\2 & 0 \\\end{matrix} \right]\), \(B=\left[ \begin{matrix}2 \\3 \\\end{matrix} \right]\).
2. \(A=\left[ \begin{align}& \begin{matrix}0 & 0 & 5 \\\end{matrix} \\&\begin{matrix}1 & -1 & 0 \\\end{matrix} \\\end{align} \right]\), \(B=\left[ \begin{matrix}1 & 2 \\3 & -1 \\4 & 5 \\\end{matrix} \right]\).
PROPERTIES OF MATRIX ADDITION: Matrix addition is commutative i.e., if A and B are two m x n matrices, then A + B = B + A.
Existence of Inverse: For every matrix A = [aij] mxn there exists a matrix [- aij] mxn, denoted by -A, such that A + (-A) = 0 = (-A) + A.
MULTIPLICATION OF A MATRIX BY A SCALAR
DEFINITION: Let A = [aij] be an m x n matrix and k be any number called a scalar. Then the matrix obtained by multiplying every element of A by k is called the scalar multiple of A by k and is denoted by kA.
Thus, kA = [kaij] mxn.
SUBTRACTION OF MATRICES
DEFINITION: For two matrices A and B of the same order, we define A – B = A + (- B).
ILLUSTRATION 2: If \(A=\left[ \begin{matrix}2 & 3 & 4 \\0 & 4 & 6 \\5 & 8 & 9\\\end{matrix} \right]\), \(B=\left[ \begin{matrix}3 & 0 & 5 \\5 & 3 & 2 \\0 & 4 & 7 \\\end{matrix} \right]\) find 3A – 2B.
SOLUTION: We have.
3A – 2B = 3A + (-2) B
\(=\left[ \begin{matrix}6 & 9 & 12 \\0 & 12 & 18 \\15 & 24 & 27 \\\end{matrix} \right]+\left[ \begin{matrix}-6 & 0 & -10 \\-10 & -6 & -4 \\0 & -8 & -14 \\\end{matrix} \right]\).
\(=\left[ \begin{matrix}0 & 9 & 12 \\-10 & 6 & 14 \\15 & 16 & 13 \\\end{matrix} \right]\).
MULTIPLICATION OF MATRICES: Two matrices A and B are conformable for the product AB if the number of column in A (pre – multiplier) is same as the number of tows in B (post-multiplier).
ILLUSTRATION 2: Let \(A=\left[ \begin{align}& \begin{matrix}1 & -2 & 3 \\\end{matrix} \\& \begin{matrix}3 & 2 & -1 \\\end{matrix} \\\end{align} \right]\) and \(B=\left[ \begin{matrix}2 & 3 \\-1 & 2 \\4 & -5 \\\end{matrix} \right]\). Find AB and BA and 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 of order 2 x 2.
\(AB=\left[ \begin{matrix}1 & -2 & 3 \\3 & 2 & -1 \\\end{matrix} \right]\left[ \begin{matrix}2 & 3 \\-1 & 2 \\4 & -5 \\\end{matrix} \right]\).
\(=\left[ \begin{matrix}2+2+12 & 3-4-15 \\6-2-4 & 9+4+5 \\\end{matrix} \right]\).
\(=\left[ \begin{matrix}16 & -16 \\0 & 18 \\\end{matrix} \right]\).
Again, B is a 3 x 2 matrix and A is a 2 x 3 matrix. So, BA exists and it is of order 3 x 3.
\(\,BA=\left[ \begin{matrix}2 & 3 \\-1 & 2 \\4 & -5 \\\end{matrix} \right]\left[ \begin{matrix}1 & -2 & 3 \\3 & 2 & -1 \\\end{matrix} \right]\).
\(=\left[ \begin{matrix}2+9 & -4+6 & 6-3 \\-1+6 & 2+4 & -3-2 \\4-15 & -8-10 & 12+5 \\\end{matrix} \right]=\left[ \begin{matrix}2+9 & -4+6 & 6-3 \\-1+6 & 2+4 & -3-2 \\4-15 & -8-10 & 12+5 \\\end{matrix} \right]\).
\(=\left[ \begin{matrix}11 & 2 & 3 \\5 & 6 &-5 \\-11 & -18 & 17 \\\end{matrix} \right]=\left[ \begin{matrix}11 & 2 & 3 \\5 & 6 & -5 \\-11 & -18 & 17 \\\end{matrix} \right]\).
Now,
Hence, AB ≠ BA.
PROPERTIES OF MATRIX MULTIPLICATION:
i. Matrix multiplication is not commutative in general.
ii. Matrix multiplication is associative i.e. (AB) C=A (BC),whenever both sides are defined.
iii. Matrix multiplication is distributive over matrix addition i.e.,
A (B + C) = AB + AC,
(A + B) C = AB + AC, whenever both sides of equality are defined.
If A is an m x n matrix, then ImA = A = A In
The product of two matrices can be the null matrix while neither of them is the null matrix.
For example, if \(A=\left[ \begin{matrix}0 & 2 \\0 & 0 \\\end{matrix} \right]\) and \(B=\left[ \begin{matrix}1 & 0 \\0 & 0 \\\end{matrix} \right]\), the \(AB=\left[ \begin{matrix}0 & 0 \\0 & 0 \\\end{matrix} \right]\) while neither A nor B is the null matrix
If A is m x n matrix and 0 is a null matrix, then
1) AmxnOnxp = Omxp
2) OpxmAmxn = Opxn
TRANSPOE OF A MATRIX
DEFINITION: Let A = [aij] be an m x n matrix. Then the transpose of A, denoted by AT or A’, is an n x m matrix such that (AT)ij = aij for all I = 1, 2, …, m, j = 1, 2, …, n.
Thus, AT is obtained from A by changing its rows into columns and its columns into rows. For example, if
\(A=\left[ \begin{matrix}1 & 2 & 3 & 4 \\2 & 3 & 4 & 1 \\3 & 2 & 1 & 4 \\\end{matrix} \right]\), then\({{A}^{T}}=\left[ \begin{matrix}1 & 2 & 3 \\2 & 3 & 2 \\3 & 4 & 1 \\4 & 1 & 4 \\\end{matrix} \right]\).
PROPERTIES OF TRANSPOSE: Let A and B be two matrices. Then.
i) (AT)T= A
ii) (A + B)T= AT + BT, A and B being of the same order.
iii) (KA) T = kAT, k be any scalar (real or complex)
iv) (AB)T= BTAT, A and B being conformable for the product AB