Matrix Multiplication Practice Questions

A matrix is a set of numbers arranged in rows and columns to form a rectangular array. Multiplying a matrix by another matrix is called "matrix multiplication."

Solved Examples

Problem 1: If the matrix A = \begin{pmatrix} 18 \\ 15 \\ -21 \end{pmatrix}, find the scalar multiple −1/3A

Solution:

To find (-1/3) A, we have to multiply every element of A by (-1/3). Then

(-1/3) A= 18 x (-1/3) 15 x (-1/3) -21 x (-1/3)

= \begin{pmatrix} -6\\ -5\\ 7 \end{pmatrix}

Problem 2: Find the product of A and B: A = \begin{pmatrix} 3 & 2 & -1 \\ 4 & 2 & 0 \end{pmatrix} and B = \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 3 & 1 \end{pmatrix}

Solution:

Given A = \begin{pmatrix} 3 & 2 & -1 \\ 4 & 2 & 0 \end{pmatrix} and B = \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 3 & 1 \end{pmatrix}

Product Matrix AB = \begin{pmatrix} 3 \cdot 0 + 2 \cdot 1 + (-1) \cdot 3 & 3 \cdot 1 + 2 \cdot 2 + (-1) \cdot 1 \\ 4 \cdot 0 + 2 \cdot 1 + 0 \cdot 3 & 4 \cdot 1 + 2 \cdot 2 + 0 \cdot 1 \end{pmatrix}

AB = \begin{pmatrix} -1 & 6 \\ 2 & 8 \end{pmatrix}

Problem 3: Find the product of the following matrices:

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{pmatrix}

B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{pmatrix}

Solution:

Given

A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{pmatrix}

B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{pmatrix}

Then,

A * B = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 2 & 1 \\ 1 & 2 & 5 \end{pmatrix} * \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 2 & 1 \end{pmatrix}

= \begin{pmatrix} 1 + 0 + 6 & 0 + 2 + 3 \\ 0 + 0 + 2 & 0 + 2 + 1 \\ 1 + 0 + 10 & 0 + 2 + 5 \end{pmatrix}

= \begin{pmatrix} 7 & 5 \\ 2 & 3 \\ 11 & 7 \end{pmatrix}

Problem 4: Calculate AB and BA. If A= \begin{bmatrix} 2 & 3\\ 1 & 4 \end{bmatrix}, \qquad B= \begin{bmatrix} 1 & 5\\ -2 & 3 \end{bmatrix}

Solution:

Find AB = \begin{bmatrix} 2(1)+1(5) & 2(-2)+1(3)\\ 3(1)+4(5) & 3(-2)+4(3) \end{bmatrix}

= \begin{bmatrix} 7 & -1\\ 23 & 6 \end{bmatrix}

Therefore: AB = = \begin{bmatrix} 7 & -1\\ 23 & 6 \end{bmatrix}

Find BA = \begin{bmatrix} 1 & 5\\ -2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3\\ 1 & 4 \end{bmatrix}

= \begin{bmatrix} 1(2)+(-2)(3) & 1(1)+(-2)(4)\\ 5(2)+3(3) & 5(1)+3(4) \end{bmatrix}

\begin{bmatrix} -4 & -7\\ 19 & 17 \end{bmatrix}

Thus, AB≠BA

This demonstrates an important property:

Matrix multiplication is generally not commutative.

Problem 5: Solve: A= \begin{bmatrix} x & 3\\ 2 & y \end{bmatrix}, \qquad B= \begin{bmatrix} 1 & 2\\ 4 & -1 \end{bmatrix} find AB.

Solution:

AB= \begin{bmatrix} x(1)+2(2) & x(4)+2(-1)\\ 3(1)+y(2) & 3(4)+y(-1) \end{bmatrix}

Therefore, AB= \begin{bmatrix} x+4 & 4x-2\\ 3+2y & 12-y \end{bmatrix}

Problem 6: Solve A= \begin{bmatrix} 1 & 0\\ 2 & 3 \end{bmatrix}, \qquad B= \begin{bmatrix} 2 & 1\\ 4 & 0 \end{bmatrix}, \qquad C= \begin{bmatrix} 1 & 2\\ 3 & 1 \end{bmatrix}

Solution:

AB= \begin{bmatrix} 1 & 0\\ 2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 4\\ 1 & 0 \end{bmatrix}

= \begin{bmatrix} 10 & 1\\ 12 & 0 \end{bmatrix}

Now multiply by C:

ABC= \begin{bmatrix} 10 & 1\\ 12 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 3 & 1 \end{bmatrix}

\begin{bmatrix} 10(1)+1(2) & 10(3)+1(1)\\ 12(1)+0(2) & 12(3)+0(1) \end{bmatrix}

\begin{bmatrix} 12 & 12\\ 31 & 36 \end{bmatrix}

Therefore, ABC= \begin{bmatrix} 12 & 12\\ 31 & 36 \end{bmatrix}

Practice Questions

Q1. Given matrices: A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}. Find AB

Q2. Given matrices: C = \begin{pmatrix} 5 & -1 \\ 2 & 3 \end{pmatrix} and D = \begin{pmatrix} 0 & 4 \\ -2 & 1 \end{pmatrix}. Find CD.

Q3. Given matrices: E = \begin{pmatrix} 3 & 0 & 2 \\ 1 & 4 & 5 \end{pmatrix} and F = \begin{pmatrix} 2 & 3 \\ 0 & 1 \\ 1 & 4 \end{pmatrix}. Find EF.

Q4. Given matrices: G = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} and H = \begin{pmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{pmatrix}.​​ Find GH.

Q5. Given matrices: I = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} and J = \begin{pmatrix} 1 & 3 \\ 5 & 7 \end{pmatrix}. Find IJ.

Q6. Given matrices: M = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \end{pmatrix} and N = \begin{pmatrix} 4 & 1 \\ 2 & 2 \\ 0 & 3 \end{pmatrix}. Find MN.