Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
To multiply two matrices, the number of columns in the first matrix must be equal to the number of rows in the second matrix.
Consider the multiplications of 3×3 and 3×2 matrices.
(Order of left hand matrix) x (order of right hand matrix) -> (order of product matrix).
(3 × 3 ) x (3 × 2 ) -> (3 × 2 )
The product AB can be found if the number of columns of matrix A is equal to the number of rows of matrix B. If the order of matrix A is m x n and B is n x p then the order of AB is m x p .
Question 1 :
Find the order of the product matrix AB if

Solution :
(i) Order of A is 3 x 3, order of B is 3 x 3.
The order of the matrix AB is 3 x 3.
(ii) Order of A is 4 x 3, order of B is 3 x 2.
The order of the matrix AB is 4 x 2.
(iii) Order of A is 4 x 2, order of B is 2 x 2.
The order of the matrix AB is 4 x 2.
(iv) Order of A is 4 x 5, order of B is 5 x 1.
The order of the matrix AB is 4 x 1.
(v) Order of A is 1 x 1, order of B is 1 x 3.
The order of the matrix AB is 1 x 3.
Question 2 :
If A is of order p x q and B is of order q x r what is the order of AB and BA?
Solution :
(p x q) (q x r) = p x r
The order of matrix AB is p x r.
(q x r) (p x q) = p x r
The product of matrices B and A is not possible.
Hence it is not defined.
Question 3 :
A has ‘a’ rows and ‘a + 3 ’ columns. B has ‘b’ rows and ‘17–b’ columns, and if both products AB and BA exist, find a, b?
Solution :
Number of rows of A = a
Number of columns of A = a + 3
Number of rows of B = b
Number of columns of A = 17 - b
Since the product of matrices A and B is possible,
[a x (a + 3)] [b x (17 - b)]
a + 3 = b
a - b = -3 -----(1)
Since the product of matrices B and A is possible,
[b x (17 - b)] [a x (a + 3)]
17 - b = a
a + b = 17 ----(2)
(1) + (2)
2a = 14
a = 7
By applying the value of a in (1), we get
7 - b = -3
-b = -3 - 7
-b = -10
b = 10
Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 05, 25 04:04 AM
Dec 03, 25 07:02 AM
Dec 02, 25 09:27 AM