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 =
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:
Solution:
Given
A = \begin{pmatrix} 3 & 2 & -1 \\ 4 & 2 & 0 \end{pmatrix} andB = \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:
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
Solution:
AB = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}
=\begin{bmatrix}2(1)+3(-2)& 2(5)+3(3)\\1(1)+4(-2) &1(5)+4(3)\end{bmatrix}
=\begin{bmatrix} -4 & 19\\ -7 & 17 \end{bmatrix} Now,
BA=\begin{bmatrix}1 & 5\\-2 & 3\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 4\end{bmatrix}
=\begin{bmatrix} 1(2)+5(1) & 1(3)+ 5(4)\\ (-2)(2)+3(1) & (-2)(3) +3(4) \end{bmatrix}
=\begin{bmatrix} 7 & 23\\ -1 & 6 \end{bmatrix} Thus, AB ≠BA
This demonstrates that matrix multiplication is generally not commutative.
Problem 5: Solve:
Solution:
AB=\begin{bmatrix}x & 3\\2 & y\end{bmatrix}\begin{bmatrix}1 & 2\\4 & -1\end{bmatrix}
=\begin{bmatrix} x(1)+3(4) & x(4)+2(-1)\\ 3(1)+y(2) & 3(4)+y(-1) \end{bmatrix}
=\begin{bmatrix} x+12 & 2x-3\\ 2+4y & 4-y \end{bmatrix} Therefore ,
AB =
=\begin{bmatrix} x+12 & 2x-3\\ 2+4y & 4-y \end{bmatrix}
Problem 6: Solve
Solution:
AB= \begin{bmatrix} 1 & 0\\ 2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1\\ 4 & 0 \end{bmatrix}
=\begin{bmatrix} 1(2)+0(4) & 1(1)+0(0)\\ 2(2)+3(4) & 2(1)+3(0) \end{bmatrix}
=\begin{bmatrix} 2 & 1\\ 16 & 2 \end{bmatrix} Now,
ABC=\begin{bmatrix}2 & 1\\16 & 2\end{bmatrix}\begin{bmatrix}1 & 2\\3 & 1\end{bmatrix}
=\begin{bmatrix} 2(1)+1(3) & 2(2)+1(1)\\ 16(1)+2(3) & 16(2)+2(1) \end{bmatrix}
=\begin{bmatrix} 5 & 5\\ 22 & 34 \end{bmatrix} Therefore ,
ABC
=\begin{bmatrix} 5 & 5\\ 22 & 34 \end{bmatrix}
Practice Questions
Q1. Given matrices: A =
Q2. Given matrices: C =
Q3. Given matrices: E =
Q4. Given matrices: G =
Q5. Given matrices: I =
Q6. Given matrices: M =