Matrix multiplication questions with solutions are given here for practice. These matrix multiplication questions will not only help students with the preparation for board exams but also with many competitive exams as well.

Also check: Important questions for class 12 Matrices.

In linear algebra, a matrix is an arrangement of elements in the form of rows and columns. An m × n matrix has m rows and n columns. If a matrix has an equal number of rows and columns, then the matrix is called a square matrix. When we perform multiplication operations with matrices, it is called matrix multiplication.

Matrix Multiplication:

If A is an m × n matrix and B is an n × p matrix, such that C = A × B is a product matrix of order m × p, then

\begin{array}{l} A_{m \times n} = [\begin{array}{c} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ \dots & \dots & \dots & \dots \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{array}], B_{n \times p} = [\begin{array}{c} b_{11} & b_{12} & \dots & b_{1 p} \\ b_{21} & b_{22} & \dots & b_{2 p} \\ \dots & \dots & \dots & \dots \\ b_{n 1} & b_{n 2} & \dots & b_{n p} \end{array}] \end{array}

\begin{array}{l} a n d C_{m \times p} = [\begin{array}{c} c_{11} & c_{12} & \dots & c_{1 p} \\ c_{21} & c_{22} & \dots & c_{2 p} \\ \dots & \dots & \dots & \dots \\ c_{m 1} & c_{m 2} & \dots & c_{m p} \end{array}] \end{array}

If c_ij ∈ C then

\begin{array}{l} c_{i j} = a_{i 1} b_{1 j} + a_{i 2} b_{2 j} + \dots + a_{i n} b_{n j} = \sum_{k = 1}^{n} a_{i k} b_{k j} \end{array}

Important Condition for Matrix Multiplication:

Two matrices are multiplication conformable if and only if the number of columns of the first matrix is equal to the rows of the second matrix.

Know more about the properties of matrix multiplication.

Matrix Multiplication Questions with Solutions

Let us solve some questions to practise matrix multiplications.

Question 1:

Find the product of the following matrices:

\begin{array}{l} A = [\begin{array}{c} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{array}] a n d B = [\begin{array}{c} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{array}] \end{array}

Solution:

Given,

\begin{array}{l} A = [\begin{array}{c} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{array}] a n d B = [\begin{array}{c} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{array}] \end{array}

Then,

\begin{array}{l} A \times B = [\begin{array}{c} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{array}] \times [\begin{array}{c} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} 1 + 0 + 6 & 0 + 2 + 3 \\ 0 + 0 + 2 & 0 + 2 + 1 \\ 1 + 0 + 10 & 0 + 2 + 5 \end{array}] = [\begin{array}{c} 7 & 5 \\ 2 & 3 \\ 11 & 7 \end{array}] \end{array}

\begin{array}{l} ∴ A \times B = [\begin{array}{c} 7 & 5 \\ 2 & 3 \\ 11 & 7 \end{array}] \end{array}

Question 2:

Prove that for the matrices A and B, (A + B)² ≠ A² + 2AB + B² where

\begin{array}{l} A = [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] a n d B = [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] \end{array}

Solution:

Given,

\begin{array}{l} A = [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] a n d B = [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] \end{array}

\begin{array}{l} A + B = [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] + [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{c} 1 & 1 \\ 1 & 5 \end{array}] \end{array}

\begin{array}{l} (A + B)^{2} = [\begin{array}{c} 1 & 1 \\ 1 & 5 \end{array}] \times [\begin{array}{c} 1 & 1 \\ 1 & 5 \end{array}] = [\begin{array}{c} 2 & 6 \\ 6 & 26 \end{array}] \dots . (i) \end{array}

\begin{array}{l} A^{2} = [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] \times [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] = [\begin{array}{c} 1 & 0 \\ 8 & 9 \end{array}] \end{array}

\begin{array}{l} B^{2} = [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] \times [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{c} 0 & 3 \\ - 3 & 3 \end{array}] \end{array}

\begin{array}{l} A \times B = [\begin{array}{c} 1 & 0 \\ 2 & 3 \end{array}] \times [\begin{array}{c} 1 & 1 \\ - 1 & 2 \end{array}] = [\begin{array}{c} 1 & 1 \\ - 1 & 8 \end{array}] \end{array}

\begin{array}{l} 2 A B = 2 [\begin{array}{c} 1 & 1 \\ - 1 & 8 \end{array}] = [\begin{array}{c} 2 & 2 \\ - 2 & 16 \end{array}] \end{array}

\begin{array}{l} A^{2} + 2 A B + B^{2} = [\begin{array}{c} 1 & 0 \\ 8 & 9 \end{array}] + [\begin{array}{c} 0 & 3 \\ - 3 & 3 \end{array}] + [\begin{array}{c} 2 & 2 \\ - 2 & 16 \end{array}] = [\begin{array}{c} 3 & 5 \\ 3 & 28 \end{array}] \dots . (i i) \end{array}

From (i) and (ii), it is proved that (A + B)² ≠ A² + 2AB + B².

Question 3:

For given matrices A and B, find AB and BA, also prove that AB ≠ BA.

\begin{array}{l} A = [\begin{array}{c} 4 & 2 & - 1 \\ 3 & 2 & 0 \end{array}] a n d B = [\begin{array}{c} 0 & 1 \\ 2 & 3 \\ 7 & 4 \end{array}] \end{array}

Solution:

Given,

\begin{array}{l} A = [\begin{array}{c} 4 & 2 & - 1 \\ 3 & 2 & 0 \end{array}] a n d B = [\begin{array}{c} 0 & 1 \\ 2 & 3 \\ 7 & 4 \end{array}] \end{array}

\begin{array}{l} A B = [\begin{array}{c} 4 & 2 & - 1 \\ 3 & 2 & 0 \end{array}] [\begin{array}{c} 0 & 1 \\ 2 & 3 \\ 7 & 4 \end{array}] = [\begin{array}{c} - 3 & 6 \\ 4 & 9 \end{array}] \dots (i) \end{array}

\begin{array}{l} B A = [\begin{array}{c} 0 & 1 \\ 2 & 3 \\ 7 & 4 \end{array}] [\begin{array}{c} 4 & 2 & - 1 \\ 3 & 2 & 0 \end{array}] = [\begin{array}{c} 3 & 2 & 0 \\ 17 & 10 & - 2 \\ 40 & 22 & - 7 \end{array}] \dots . (i i) \end{array}

From (i) and (ii) AB ≠ BA.

Also Read:

Question 4:

Let X, Y, Z, W and S be matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Find the values of n, k, and p so that SY + WY is well defined.

Solution:

Order of S = p × k

Order of Y = 3 × k

If the matrices S and Y are multiplication conformable, then k = 3.

Order of WY = n × k

Order of SY = p × k

If the matrices SY and WY are addition conformable, then p = n.

Question 5:

Verify the associative property for multiplication of matrices:

\begin{array}{l} A = [\begin{array}{c} 1 & 2 \\ - 2 & 3 \end{array}], B = [\begin{array}{c} 2 & 1 \\ 2 & 3 \end{array}] a n d C = [\begin{array}{c} - 3 & 1 \\ 2 & 0 \end{array}] \end{array}

Solution:

Given,

\begin{array}{l} A = [\begin{array}{c} 1 & 2 \\ - 2 & 3 \end{array}], B = [\begin{array}{c} 2 & 1 \\ 2 & 3 \end{array}] a n d C = [\begin{array}{c} - 3 & 1 \\ 2 & 0 \end{array}] \end{array}

We need to prove that A(BC) = (AB)C.

Then

\begin{array}{l} B C = [\begin{array}{c} 2 & 1 \\ 2 & 3 \end{array}] [\begin{array}{c} - 3 & 1 \\ 2 & 0 \end{array}] = [\begin{array}{c} - 4 & 2 \\ 0 & 2 \end{array}] \end{array}

\begin{array}{l} A (B C) = [\begin{array}{c} 1 & 2 \\ - 2 & 3 \end{array}] [\begin{array}{c} - 4 & 2 \\ 0 & 2 \end{array}] = [\begin{array}{c} - 4 & 6 \\ 8 & 2 \end{array}] \dots (i) \end{array}

\begin{array}{l} A B = [\begin{array}{c} 1 & 2 \\ - 2 & 3 \end{array}] [\begin{array}{c} 2 & 1 \\ 2 & 3 \end{array}] = [\begin{array}{c} 6 & 7 \\ 2 & 7 \end{array}] \end{array}

\begin{array}{l} (A B) C = [\begin{array}{c} 6 & 7 \\ 2 & 7 \end{array}] [\begin{array}{c} - 3 & 1 \\ 2 & 0 \end{array}] = [\begin{array}{c} - 4 & 6 \\ 8 & 2 \end{array}] \dots . (i i) \end{array}

From (i) and (ii) we get that A(BC) = (AB)C.

Question 6:

Show that the product of the two matrices

\begin{array}{l} [\begin{array}{c} c o s^{2} θ & c o s θ s i n θ \\ c o s θ s i n θ & s i n^{2} θ \end{array}] a n d [\begin{array}{c} c o s^{2} ϕ & c o s ϕ s i n ϕ \\ c o s ϕ s i n ϕ & s i n^{2} ϕ \end{array}] \end{array}

It is a null matrix when the difference between both angles is an odd multiple of 𝜋/2.

Solution:

The product of both the matrix is given by

\begin{array}{l} [\begin{array}{c} c o s^{2} θ & c o s θ s i n θ \\ c o s θ s i n θ & s i n^{2} θ \end{array}] [\begin{array}{c} c o s^{2} ϕ & c o s ϕ s i n ϕ \\ c o s ϕ s i n ϕ & s i n^{2} ϕ \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} c o s^{2} θ c o s^{2} ϕ + c o s θ c o s ϕ s i n θ s i n ϕ & c o s^{2} θ c o s ϕ s i n ϕ + c o s θ s i n θ s i n^{2} ϕ \\ c o s^{2} ϕ c o s θ s i n θ + c o s ϕ s i n ϕ s i n^{2} θ & s i n^{2} θ s i n^{2} ϕ + c o s θ c o s ϕ s i n θ s i n ϕ \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} c o s θ c o s ϕ c o s (θ - ϕ) & c o s θ s i n ϕ c o s (θ - ϕ) \\ s i n θ c o s ϕ c o s (θ - ϕ) & s i n θ s i n ϕ c o s (θ - ϕ) \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}] [a s θ - ϕ = (2 n + 1) \frac{π}{2}, n ϵ Z, t h e n c o s [(2 n + 1) \frac{π}{2}] = 0] \end{array}

Question 7:

Let the given matrices A, B and C be

\begin{array}{l} A = [\begin{array}{c} 1 & 0 \\ 2 & 4 \end{array}], B = [\begin{array}{c} 2 & 4 \\ 1 & 0 \end{array}] a n d C = [\begin{array}{c} - 1 & 5 \\ 1 & - 1 \end{array}] \end{array}

Find a matrix D such that CD – AB = 0

Solution:

Given,

\begin{array}{l} A = [\begin{array}{c} 1 & 0 \\ 2 & 4 \end{array}], B = [\begin{array}{c} 2 & 4 \\ 1 & 0 \end{array}] a n d C = [\begin{array}{c} - 1 & 5 \\ 1 & - 1 \end{array}] \end{array}

\begin{array}{l} L e t, D = [\begin{array}{c} a & b \\ c & d \end{array}] \end{array}

Given, CD – AB = 0 ⇒ CD = AB

\begin{array}{l} A B = [\begin{array}{c} 1 & 0 \\ 2 & 4 \end{array}] [\begin{array}{c} 2 & 4 \\ 1 & 0 \end{array}] = [\begin{array}{c} 2 & 4 \\ 8 & 8 \end{array}] \end{array}

\begin{array}{l} C D = [\begin{array}{c} - 1 & 5 \\ 1 & - 1 \end{array}] [\begin{array}{c} a & b \\ c & d \end{array}] = [\begin{array}{c} - a + 5 c & - b + 5 d \\ - a + c & - b + d \end{array}] \end{array}

Then,

\begin{array}{l} [\begin{array}{c} - a + 5 c & - b + 5 d \\ - a + c & - b + d \end{array}] = [\begin{array}{c} 2 & 4 \\ 8 & 8 \end{array}] \end{array}

We get the following equations,

– a + 5c = 2 ….(i)

– a + c = 8 ….(ii)

– b + 5d = 4 ….(iii)

– b + d = 8 ….(iv)

Solving (i) and (ii) we get, a = – 19/2 and c = –3/2

Solving (iii) and (iv) we get, b = –9 and d = –1.

Therefore, we get the matrix

\begin{array}{l} D = [\begin{array}{c} \frac{- 19}{2} & - 9 \\ \frac{- 3}{2} & - 1 \end{array}] \end{array}

Also Read:

Question 8:

Show that A² – 3I = 2A where

\begin{array}{l} A = [\begin{array}{c} 1 & 2 \\ 2 & 1 \end{array}] \end{array}

Solution:

\begin{array}{l} N o w, A^{2} = A . A = [\begin{array}{c} 1 & 2 \\ 2 & 1 \end{array}] . [\begin{array}{c} 1 & 2 \\ 2 & 1 \end{array}] = [\begin{array}{c} 5 & 4 \\ 4 & 5 \end{array}] \end{array}

\begin{array}{l} 3 I = 3 [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}] = [\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}] \end{array}

\begin{array}{l} A^{2} - 3 I = [\begin{array}{c} 5 & 4 \\ 4 & 5 \end{array}] [\begin{array}{c} 3 & 0 \\ 0 & 3 \end{array}] = [\begin{array}{c} 2 & 4 \\ 4 & 2 \end{array}] \end{array}

\begin{array}{l} A l s o, 2 A = 2 [\begin{array}{c} 1 & 2 \\ 2 & 1 \end{array}] = [\begin{array}{c} 2 & 4 \\ 4 & 2 \end{array}] \end{array}

Hence, A² – 3I = 2A.

Question 9:

Let f(x) = 2x² – 3x, find det [f(A)] if

\begin{array}{l} A = [\begin{array}{c} - 2 & 1 \\ 0 & 3 \end{array}] \end{array}

Solution:

Given, f(x) = 2x² – 3x, then f(A) = 2A² – 3A

\begin{array}{l} 2 A^{2} - 3 A = 2 [\begin{array}{c} - 2 & 1 \\ 0 & 3 \end{array}] [\begin{array}{c} - 2 & 1 \\ 0 & 3 \end{array}] - 3 [\begin{array}{c} - 2 & 1 \\ 0 & 3 \end{array}] \end{array}

\begin{array}{l} = 2 [\begin{array}{c} 4 & 1 \\ 0 & 9 \end{array}] - [\begin{array}{c} - 6 & 3 \\ 0 & 9 \end{array}] \end{array}

\begin{array}{l} = [\begin{array}{c} 8 & 2 \\ 0 & 18 \end{array}] - [\begin{array}{c} - 6 & 3 \\ 0 & 9 \end{array}] = [\begin{array}{c} 14 & - 1 \\ 0 & 9 \end{array}] \end{array}

\begin{array}{l} \Rightarrow f (A) = [\begin{array}{c} 14 & - 1 \\ 0 & 9 \end{array}] \end{array}

\begin{array}{l} | f (A) | = | \begin{array}{c} 14 & - 1 \\ 0 & 9 \end{array} | = 14 \times 9 - 0 = 126 \end{array}

Question 10:

Show that f(x).f(y) = f(x + y) where

\begin{array}{l} f (θ) = [\begin{array}{c} c o s θ & - s i n θ & 0 \\ s i n θ & c o s θ & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

Solution:

We have

\begin{array}{l} f (x) = [\begin{array}{c} c o s x & - s i n x & 0 \\ s i n x & c o s x & 0 \\ 0 & 0 & 1 \end{array}], f (y) = [\begin{array}{c} c o s y & - s i n y & 0 \\ s i n y & c o s y & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

\begin{array}{l} a n d f (x + y) = [\begin{array}{c} c o s (x + y) & - s i n (x + y) & 0 \\ s i n (x + y) & c o s (x + y) & 0 \\ 0 & 0 & 1 \end{array}] \end{array}

\begin{array}{l} f (x) . f (y) = [\begin{array}{c} c o s x & - s i n x & 0 \\ s i n x & c o s x & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{c} c o s y & - s i n y & 0 \\ s i n y & c o s y & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{array}{c} c o s (x + y) & - s i n (x + y) & 0 \\ s i n (x + y) & c o s (x + y) & 0 \\ 0 & 0 & 1 \end{array}] = f (x + y) \end{array}

Practice Questions on Matrix Multiplication

1. Find the product of the following matrices:

\begin{array}{l} A = [\begin{array}{c} 1 & - 2 & 3 \\ 3 & 2 & - 1 \end{array}] a n d B = [\begin{array}{c} 2 & 3 \\ - 1 & 2 \\ 4 & - 5 \end{array}] \end{array}

2. Let X, Y, Z, W and S are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Find the order of matrix 7X – 5Z, if n = p.

3. Verify the distributive property for multiplication over addition for the matrices:

\begin{array}{l} A = [\begin{array}{c} 1 & 2 \\ - 2 & 3 \end{array}], B = [\begin{array}{c} 2 & 1 \\ 2 & 3 \end{array}] a n d C = [\begin{array}{c} - 3 & 1 \\ 2 & 0 \end{array}] \end{array}

4. For the given matrices A and B, verify (A + B)² = A² + B², where

\begin{array}{l} A = [\begin{array}{c} 7 & 2 \\ 0 & 1 \end{array}] a n d B = [\begin{array}{c} 2 & 3 \\ 0 & - 1 \end{array}] \end{array}

Learn more about Matrices and other related topics in a fun and enjoyable way. Download BYJU’S – The Learning App today.

Learn about various mathematical concepts in a simple manner with detailed information, along with step-by-step solutions to all questions, only at BYJU’S. Download BYJU’S – The Learning App to get personalised videos.

Related Articles
Permutation and Combination Questions	Differentiation Questions
Functions Questions	Trigonometry Questions

Matrix Multiplication Questions with Solutions

Recommended Videos

Related Articles

Practice Questions on Matrix Multiplication