How to Multiply Matrices

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Matrices Operations

The basic operations on a matrix are addition, subtraction, and multiplication. For matrices to be added or subtracted, they must be in the same order. For matrices to be multiplied, the number of columns in the first Matrix must equal the number of rows in the second Matrix.

How to Multiply Matrixes

If $A$ and $B$ be any two matrices, then their product $A \times B$ will be defined only when the number of columns in $A$ is equal to the number of rows in $B$ .

If $A \ = \ [a_{ij}]_{m \times n}$ and $B \ = \ [b_{ij}]_{n \times p}$ then, their final product will be $A \times B \ = \ C \ = \ [c_{ij}]_{m \times p}$ :

$C_{ij} \ = \ \sum_{t=1}^n a_{it}b_{tj}$

Consider $A \ = \ \begin{bmatrix} a_{11} & a_{12} \\\ a_{21} & a_{22} \end{bmatrix}$ , $B \ = \ \begin{bmatrix} b_{11} & b_{12} \\\ b_{21} & b_{22} \end{bmatrix}$

$⇒ \ A \times B \ = \ C \ = \ \begin{bmatrix} a_{11}b_{11} \ + \ a_{12}b_{21} & a_{11}b_{12} \ + \ a_{12}b_{22} \\\ a_{21}b_{11} \ + \ a_{22}b_{21} & a_{21}b_{12} \ + \ a_{22}b_{22} \end{bmatrix}$

Matrix Multiplication Properties

In general, matrix multiplication is not commutative: $AB \ ≠ \ BA$

Multiplication of matrices is associative, which means that: $(AB)C \ = \ A(BC)$

Multiplication of matrices is distributive over the addition of matrices, which means that $A(B \ + \ C) \ = \ AB \ + \ AC$ and $(A \ + \ B)C \ = \ AC \ + \ BC$ .

Is $A$ an $m \times n$ matrix? If so, then $I_mA \ = \ A \ = \ AI_n$ .

The product of two matrices can be a null matrix even if neither of the two matrices is null. This means that if $AB \ = \ 0$ , it is not necessary that $A \ = \ 0$ or $B \ = \ 0$ .

If $A$ is an $m \times n$ matrix and $O$ is a null matrix, then the product of a matrix and a null matrix is always a null matrix: $A_{m \times n}O_{n \times p} \ = \ O_{m \times p}$

7- If $AB \ = \ 0$ , it does not necessarily follow that $A$ or $B$ are zero, and a zero matrix could result from multiplying two non-zero matrices.

If $AB \ = \ AC$ , $B \ ≠ \ C$ (Cancellation Law is not applicable).

$tr(AB) \ = \ tr(BA)$

10- For every square Matrix, there is a multiplicative identity such that $AI \ = \ IA \ = A$

Free printable Worksheets

Exercises for Matrix Multiplication

1) Find the answer: $\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix}$

2) Find the answer: $\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix}$

3) Find the answer: $\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix}$

4) Find the answer: $\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix}$

5) Find the answer: $\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix}$

6) Find the answer: $\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix}$

7) Find the answer: $\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix}$

8) Find the answer: $\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix}$

9) Find the answer: $\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix}$

10) Find the answer: $\begin{bmatrix} 5 & 4 & 9 \end{bmatrix} \times \begin{bmatrix} 7 \\\ -6 \\\ 3 \end{bmatrix}$

Answers

1) Find the answer: $\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix}$

$\color{red}{\begin{bmatrix} -6 & 2 \\\ 3 & -5 \end{bmatrix} \times \begin{bmatrix} 1 & 3 \\\ 0 & -1 \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} (-6 \times 1) \ + \ (2 \times 0) & (-6 \times 3) \ + \ (2 \times -1) \\\ (3 \times 1) \ + \ (-5 \times 0) & (3 \times 3) \ + \ (-5 \times -1) \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} -6 & -20 \\\ 3 & 14 \end{bmatrix}}$

2) Find the answer: $\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 4 & -8 \\\ -2 & 3 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \\\ 1 & 3 \end{bmatrix} \ = }$ $\color{red}{ \begin{bmatrix} (4 \times 2) \ + \ (-8 \times 1) & (4 \times 0) \ + \ (-8 \times -3) \\\ (-2 \times 2) \ + \ (3 \times 1) & (-2 \times 0) \ + \ (3 \times 3) \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} 0 & -24 \\\ -1 & 9 \end{bmatrix}}$

3) Find the answer: $\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 7 \\\ 3 \\\ 4 \end{bmatrix} \times \begin{bmatrix} -2 & 1 \end{bmatrix} \ = \begin{bmatrix} (7 \times (-2)) & (7 \times 1) \\\ (3 \times (-2)) & (3 \times 1) \\\ (4 \times (-2)) & (4 \times 1) \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} -14 & 7 \\\ -6 & 3 \\\ -8 & 4 \end{bmatrix}}$

4) Find the answer: $\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 3 \\\ 9 \end{bmatrix} \times \begin{bmatrix} 7 & 3 & -2 \end{bmatrix} \ = }$ $\color{red}{ \begin{bmatrix} (3 \times 7) & (3 \times 3) & (3 \times (-2)) \\\ (9 \times 7) & (9 \times 3) & (9 \times (-2)) \end{bmatrix} \ =}$ $\color{red}{\begin{bmatrix} 21 & 9 & -6 \\\ 63 & 27 & -18 \end{bmatrix}}$

5) Find the answer: $\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix} \ =}$ $\color{red}{\begin{bmatrix} (7 \times 3) \ + \ (-2 \times 9) \end{bmatrix} \ = \ \begin{bmatrix} 3 \end{bmatrix}}$

6) Find the answer: $\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix}$

$\color{red}{\begin{bmatrix} -4 & 1 \\\ -3 & 5 \end{bmatrix} \times \begin{bmatrix} 7 & 9 \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} (-4 \times 7) \ + \ (1 \times 9) & (-3 \times 7) \ + \ (5 \times 9) \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} -19 & 24 \end{bmatrix}}$

7) Find the answer: $\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 4 \\\ -2 \\\ 3 \end{bmatrix} \times \begin{bmatrix} 7 & 5 \end{bmatrix} \ = \ \begin{bmatrix} (4 \times 7) & (4 \times 5) \\\ ((-2) \times 7) & ((-2) \times 5) \\\ (3 \times 7) & (3 \times 5) \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} 28 & 20 \\\ -14 & -10 \\\ 21 & 15 \end{bmatrix}}$

8) Find the answer: $\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix}$

$\color{red}{\begin{bmatrix} -9 \\\ 8 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & -4 \end{bmatrix} \ =}$ $\color{red}{ \begin{bmatrix} (-9 \times 3) & (-9 \times (-2)) & (-9 \times (-4)) \\\ (8 \times 3) & (8 \times (-2)) & (8 \times (-4)) \end{bmatrix} \ = }$ $\color{red}{\begin{bmatrix} -27 & 18 & 36 \\\ 24 & -16 & -32 \end{bmatrix}}$

9) Find the answer: $\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 5 & -7 \end{bmatrix} \times \begin{bmatrix} 4 & 3 \\\ 7 & -6 \end{bmatrix} \ = }$ $\color{red}{ \begin{bmatrix} (5 \times 4) \ + \ (-7 \times 7) & (5 \times 3) \ + \ (-7 \times (-6)) \end{bmatrix} \ = }$ $\color{red}{ \begin{bmatrix} -29 & 57 \end{bmatrix}}$

10) Find the answer: $\begin{bmatrix} 5 & 4 & 9 \end{bmatrix} \times \begin{bmatrix} 7 \\\ -6 \\\ 3 \end{bmatrix}$

$\color{red}{\begin{bmatrix} 7 & -2 \end{bmatrix} \times \begin{bmatrix} 3 \\\ 9 \end{bmatrix} \ = \ \begin{bmatrix} (5 \times 7) \ + \ (4 \times (-6)) \ + \ (9 \times 3) \end{bmatrix} \ =}$ $\color{red}{ \ \begin{bmatrix} 38 \end{bmatrix}}$

How to Multiply Matrices