Matrix Scalar Multiplication

Scaling Every Matrix Entry by One Number

Matrices can be combined with other matrices, but they can also be scaled by one number. That number is called a scalar.

For example, if every entry in a measurement matrix is recorded in meters, converting the whole table to centimeters multiplies every entry by $100$ . Scalar multiplication expresses that one consistent change without rewriting each calculation separately.

Writing Scalar Multiplication Entry by Entry

Matrix scalar multiplication is the operation of multiplying every element in a matrix by a scalar number.

If we have a matrix $A$ and a scalar $k$ , then the result of the scalar multiplication of $k$ by matrix $A$ (written as $kA$ ) is a new matrix where each element is the product of the corresponding element of matrix $A$ and the scalar $k$ .

Mathematically, if matrix $A$ has an order of $m \times n$ :

A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}

Then the multiplication of matrix $A$ by scalar $k$ is:

\begin{aligned} &kA \\ &= k \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \\ &= \begin{bmatrix} ka_{11} & ka_{12} & \cdots & ka_{1n} \\ ka_{21} & ka_{22} & \cdots & ka_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ ka_{m1} & ka_{m2} & \cdots & ka_{mn} \end{bmatrix} \end{aligned}

The resulting matrix, $kA$ , will have the same order as matrix $A$ .

For a positive integer $k$ , scalar multiplication can be viewed as repeated addition. For example, $2A$ is the same as $A + A$ . Adding matrix $A$ exactly $k\text{ times}$ gives $kA$ .

\underbrace{A + A + \cdots + A}_{k \text{ times}} = kA

Matrix Scalar Multiplication Examples

The next two examples multiply every matrix entry by the same scalar.

Example 1:

Suppose we have matrix $P$ as in the following example:

P = \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix}

Determine $2P$ !

Solution:

To calculate $2P$ , we multiply each element of matrix $P$ by the scalar $2$ .

\begin{aligned} 2P &= 2 \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 \times 9 & 2 \times 7 \\ 2 \times 3 & 2 \times 1 \end{bmatrix} \\ &= \begin{bmatrix} 18 & 14 \\ 6 & 2 \end{bmatrix} \end{aligned}

The matrix above is the result of $2P$ .

Example 2:

The matrix $Q$ is given by:

Q = \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}

For the scalar $k=4$ , determine $4Q$ .

Solution:

We will multiply each element in matrix $Q$ by the scalar $4$ .

4Q = 4 \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}

Write each product with parentheses: $4(a)$ means $4\times a$ .

=\begin{bmatrix}4(-1)&4(\frac12)&4(1)\\4(4)&4(-\frac14)&4(2)\\4(-6)&4(1)&4(-4)\end{bmatrix}

The matrix above is the result of $4Q$ .

Properties of Matrix Scalar Multiplication

Let $A$ and $B$ be matrices of the same order, $h$ and $k$ be scalars, and $O$ be the zero matrix. Scalar multiplication then satisfies the following identities.

Distributive over Matrix Addition:

$k(A + B) = kA + kB$

Multiplying a scalar by the sum of two matrices is the same as summing the products of the scalar with each matrix.
Distributive over Scalar Addition:

$(h + k)A = hA + kA$

Multiplying the sum of two scalars by a matrix is the same as summing the products of each scalar with the matrix.
Associative with Scalar Multiplication:

$(hk)A = h(kA) = k(hA)$

Multiplying a matrix by the product of two scalars is the same as multiplying the first scalar by the product of the second scalar and the matrix.
Scalar Multiplication Identity:

$1A = A$

Multiplying a matrix by the scalar $1$ does not change the matrix.
Multiplication by Zero Scalar:

$0A = O$

Multiplying a matrix by the scalar $0$ results in the zero matrix ( $O$ ), which is a matrix where all elements are $0$ .
Multiplication of Zero Matrix by a Scalar:

$kO = O$

Multiplying the zero matrix by any scalar results in the zero matrix.
Multiplication by Scalar $-1$ :

$(-1)A = -A$

Multiplying a matrix by the scalar $-1$ results in the negative of the matrix.

The distributive rules expand scalar multiplication over sums, while the identity and zero rules determine the result immediately for the scalars $1$ and $0$ .

Exercises

Each problem below asks for a scalar multiple of a matrix. Multiply every entry by the scalar, including the negative entries and the zeros, and keep the shape of the matrix unchanged.

Given the following matrix, calculate $3X$ .

$X = \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}$
For the following matrix, determine $\frac{1}{10}Y$ .

$Y = \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}$
For the following matrices, show that $3(A+B) = 3A + 3B$ .
$A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$

Worked Solutions

Solution:
$3X = 3 \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}$
Solution:
$\frac{1}{10}Y = \frac{1}{10} \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}$
To show $3(A+B) = 3A + 3B$ :

First, calculate the left side of the equation, $3(A+B)$ .

$\begin{aligned} A+B &= \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 2+0 & 1+5 \\ 3+6 & 4+(-2) \end{bmatrix} \\ &= \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix} \end{aligned}$

Then,

$3(A+B) = 3 \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}$

Next, calculate the right side of the equation, $3A + 3B$ .
$3A = 3 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix}$
Then,

$\begin{aligned} &3A + 3B \\ &= \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix} + \begin{bmatrix} 0 & 15 \\ 18 & -6 \end{bmatrix} \\ &= \begin{bmatrix} 6+0 & 3+15 \\ 9+18 & 12+(-6) \end{bmatrix} \\ &= \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix} \end{aligned}$

Both calculations give the same matrix, so $3(A+B) = 3A + 3B$ .

Scaling Every Matrix Entry by One Number

Matrices can be combined with other matrices, but they can also be scaled by one number. That number is called a scalar.

Writing Scalar Multiplication Entry by Entry

Matrix scalar multiplication is the operation of multiplying every element in a matrix by a scalar number.

Mathematically, if matrix $A$ has an order of $m \times n$ :

A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}

Then the multiplication of matrix $A$ by scalar $k$ is:

\begin{aligned} &kA \\ &= k \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \\ &= \begin{bmatrix} ka_{11} & ka_{12} & \cdots & ka_{1n} \\ ka_{21} & ka_{22} & \cdots & ka_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ ka_{m1} & ka_{m2} & \cdots & ka_{mn} \end{bmatrix} \end{aligned}

The resulting matrix, $kA$ , will have the same order as matrix $A$ .

For a positive integer $k$ , scalar multiplication can be viewed as repeated addition. For example, $2A$ is the same as $A + A$ . Adding matrix $A$ exactly $k\text{ times}$ gives $kA$ .

\underbrace{A + A + \cdots + A}_{k \text{ times}} = kA

Matrix Scalar Multiplication Examples

The next two examples multiply every matrix entry by the same scalar.

Example 1:

Suppose we have matrix $P$ as in the following example:

P = \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix}

Determine $2P$ !

Solution:

To calculate $2P$ , we multiply each element of matrix $P$ by the scalar $2$ .

\begin{aligned} 2P &= 2 \begin{bmatrix} 9 & 7 \\ 3 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 2 \times 9 & 2 \times 7 \\ 2 \times 3 & 2 \times 1 \end{bmatrix} \\ &= \begin{bmatrix} 18 & 14 \\ 6 & 2 \end{bmatrix} \end{aligned}

The matrix above is the result of $2P$ .

Example 2:

The matrix $Q$ is given by:

Q = \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}

For the scalar $k=4$ , determine $4Q$ .

Solution:

We will multiply each element in matrix $Q$ by the scalar $4$ .

4Q = 4 \begin{bmatrix} -1 & \frac{1}{2} & 1 \\ 4 & -\frac{1}{4} & 2 \\ -6 & 1 & -4 \end{bmatrix}

Write each product with parentheses: $4(a)$ means $4\times a$ .

=\begin{bmatrix}4(-1)&4(\frac12)&4(1)\\4(4)&4(-\frac14)&4(2)\\4(-6)&4(1)&4(-4)\end{bmatrix}

The matrix above is the result of $4Q$ .

Properties of Matrix Scalar Multiplication

Let $A$ and $B$ be matrices of the same order, $h$ and $k$ be scalars, and $O$ be the zero matrix. Scalar multiplication then satisfies the following identities.

Distributive over Matrix Addition:

$k(A + B) = kA + kB$

Multiplying a scalar by the sum of two matrices is the same as summing the products of the scalar with each matrix.
Distributive over Scalar Addition:

$(h + k)A = hA + kA$

Multiplying the sum of two scalars by a matrix is the same as summing the products of each scalar with the matrix.
Associative with Scalar Multiplication:

$(hk)A = h(kA) = k(hA)$

Multiplying a matrix by the product of two scalars is the same as multiplying the first scalar by the product of the second scalar and the matrix.
Scalar Multiplication Identity:

$1A = A$

Multiplying a matrix by the scalar $1$ does not change the matrix.
Multiplication by Zero Scalar:

$0A = O$

Multiplying a matrix by the scalar $0$ results in the zero matrix ( $O$ ), which is a matrix where all elements are $0$ .
Multiplication of Zero Matrix by a Scalar:

$kO = O$

Multiplying the zero matrix by any scalar results in the zero matrix.
Multiplication by Scalar $-1$ :

$(-1)A = -A$

Multiplying a matrix by the scalar $-1$ results in the negative of the matrix.

The distributive rules expand scalar multiplication over sums, while the identity and zero rules determine the result immediately for the scalars $1$ and $0$ .

Exercises

Each problem below asks for a scalar multiple of a matrix. Multiply every entry by the scalar, including the negative entries and the zeros, and keep the shape of the matrix unchanged.

Given the following matrix, calculate $3X$ .

$X = \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}$
For the following matrix, determine $\frac{1}{10}Y$ .

$Y = \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}$
For the following matrices, show that $3(A+B) = 3A + 3B$ .
$A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$

Worked Solutions

Solution:
$3X = 3 \begin{bmatrix} 5 & -2 \\ 0 & 4 \\ -1 & 7 \end{bmatrix}$
Solution:
$\frac{1}{10}Y = \frac{1}{10} \begin{bmatrix} 10 & 20 \\ -30 & 0 \end{bmatrix}$
To show $3(A+B) = 3A + 3B$ :

First, calculate the left side of the equation, $3(A+B)$ .

$\begin{aligned} A+B &= \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 0 & 5 \\ 6 & -2 \end{bmatrix} \\ &= \begin{bmatrix} 2+0 & 1+5 \\ 3+6 & 4+(-2) \end{bmatrix} \\ &= \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix} \end{aligned}$

Then,

$3(A+B) = 3 \begin{bmatrix} 2 & 6 \\ 9 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix}$

Next, calculate the right side of the equation, $3A + 3B$ .
$3A = 3 \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix}$
Then,

$\begin{aligned} &3A + 3B \\ &= \begin{bmatrix} 6 & 3 \\ 9 & 12 \end{bmatrix} + \begin{bmatrix} 0 & 15 \\ 18 & -6 \end{bmatrix} \\ &= \begin{bmatrix} 6+0 & 3+15 \\ 9+18 & 12+(-6) \end{bmatrix} \\ &= \begin{bmatrix} 6 & 18 \\ 27 & 6 \end{bmatrix} \end{aligned}$

Both calculations give the same matrix, so $3(A+B) = 3A + 3B$ .