# How to Multiply a Matrix by a Scalar?

When multiplying a matrix by a scalar, the resulting matrix will always have the same dimensions as the original matrix. In this step-by-step guide, you learn more about how to multiply a matrix by a scalar.

The scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each input in the matrix is multiplied by the given scalar.

## Step by step guide to scalar multiplication of matrices

In matrix algebra, a real number is called a scalar. The scalar product of a real number, $$b$$, and a matrix $$A$$ is the matrix $$bA$$. Each element of matrix $$bA$$ is equal to $$b$$ times its corresponding element in $$A$$.

Given scalar $$\color {blue}{b}$$ and matrix $$A= \begin{bmatrix}a_{11} & a_{12} \\a_{21} & a_{22} \end{bmatrix}$$, $$\color{blue}{b}A=\begin{bmatrix}\color{blue}{b}a_{11} & \color{blue}{b}a_{12} \\\color{blue}{b}a_{21} &\color{blue}{b} a_{22} \end{bmatrix}$$

Properties of Scalar Multiplication:

Let $$A$$ and $$B$$ be $$m\times n$$ matrices. Let $$O_{m\times n}$$ be the $$m\times n$$ zero matrix and let $$p$$ and $$q$$ be scalars.

• Associative Property: $$\color{blue}{p(qA)=(pq)A}$$
• Closure Property: $$\color{blue}{pA}$$ is an $$\color{blue}{m×n}$$ matrix
• Commutative Property: $$\color{blue}{pA= Ap}$$
• Distributive Property: $$\color{blue}{(p+q)A=pA+qA}$$, $$\color{blue}{p(A+B)=pA+pB}$$
• Identity Property:$$\color{blue}{1. A=A}$$
• Multiplicative Property of $$-1$$: $$\color{blue}{(-1)A=-A}$$
• Multiplicative Property of $$0$$:$$\color{blue}{\space0. A=O_{m×n}}$$

### Scalar Multiplication of Matrices – Example 1:

If $$A=$$ $$\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}$$ , find $$3A$$.

solution:

$$3A=3$$ $$\begin{bmatrix}-5 & -5 \\-1 & 2 \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}3(-5) & 3(-5) \\3(-1) &3(2) \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}-15 & -15 \\-3& 6 \end{bmatrix}$$

### Scalar Multiplication of Matrices – Example 2:

If $$A=$$ $$\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}$$, find $$4A$$.

solution:

$$4A=$$ $$\begin{bmatrix}6 & 4 & 24 \\1 & -9 & 8 \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}4(6) & 4(4) & 4(24) \\4(1) & 4(-9) & 4(8) \end{bmatrix}$$ $$=$$ $$\begin{bmatrix}24 & 16 & 96 \\4 & -36 & 32\end{bmatrix}$$

## Exercises for Scalar Multiplication of Matrices

### Solve.

1. $$\color{blue}{A=}$$$$\color{blue}{\begin{bmatrix}0 & 2 \\-2 & -5 \end{bmatrix}}$$, $$\color{blue}{6A}$$.
2. $$\color{blue}{B=}$$$$\color{blue}{\begin{bmatrix}-5 & 0 &2 \\7& -3& 4 \\ -1& 3 & 2 \end{bmatrix}}$$, $$\color{blue}{-3B}$$.
3. $$\color{blue}{D=}$$$$\color{blue}{\begin{bmatrix}6 & -2 \\3 & 7 \end{bmatrix}}$$, $$\color{blue}{F=}$$$$\color{blue}{\begin{bmatrix}1 & -2 \\-3 & 4 \end{bmatrix}}$$, $$\color{blue}{-2D+5F}$$.
4. $$\color{blue}{A=}$$$$\color{blue}{\begin{bmatrix}-5 & 2& 0 \\7 & -4 & 3 \\ -1 & 2 & 4 \end{bmatrix}}$$, $$\color{blue}{B=}$$$$\color{blue}{\begin{bmatrix}0 & -1 & 7 \\6 & -12 & 2 \\ 9 & 5 & 1 \end{bmatrix}}$$, $$\color{blue}{4A – 3B}$$.
1. $$\color{blue}{\begin{bmatrix}0 & 12 \\-12 & -30 \end{bmatrix}}$$
2. $$\color{blue}{\begin{bmatrix}15 & 0 & -6 \\-21 & 9 &-12 \\ 3 & -9 & -6 \end{bmatrix}}$$
3. $$\color{blue}{\begin{bmatrix}-7 & -6 \\-21 & 6 \end{bmatrix}}$$
4. $$\color{blue}{\begin{bmatrix}-20 & 11 & -21 \\10 & 20 & 6 \\ -31 & -7 & 13 \end{bmatrix}}$$

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