Multiplying matrices by scalars

Learn how to find the result of a matrix multiplied by a real number.

What you should be familiar with before taking this lesson

A matrix is a rectangular arrangement of numbers into rows and columns. Each number in a matrix is referred to as a matrix element or entry.
If this is new to you, you might want to check out our intro to matrices. You should also make sure you know how to add and subtract matrices.

What you will learn in this lesson

We can multiply matrices by real numbers. This article explores how this works.

Scalars and scalar multiplication

When we work with matrices, we refer to real numbers as scalars.
The term scalar multiplication refers to the product of a real number and a matrix. In scalar multiplication, each entry in the matrix is multiplied by the given scalar.
For example, given that $A={\left[\begin{array}{rr}{10} &6 \\ 4& 3 \end{array}\right]}$, let's find 2, A.
To find 2, A, simply multiply each matrix entry by 2:
\begin{aligned}2A&=\greenD{2}\cdot{\left[\begin{array}{rr}{10} &6 \\ 4& 3 \end{array}\right]}\\\\\\ &={\left[\begin{array}{ll}{\greenD2 \cdot10} &\greenD2\cdot 6 \\ \greenD2\cdot 4& \greenD2\cdot3 \end{array}\right]}\\\\\\ &={\left[\begin{array}{rr}{20} &12 \\ 8& 6 \end{array}\right]}\end{aligned}

1) Given $B={\left[\begin{array}{rr}{-4} &-2 \\ 7& 1 \end{array}\right]}$, find minus, 3, B.
minus, 3, B, equals

To find minus, 3, B, multiply each matrix entry by minus, 3.
\begin{aligned}-3B&=\greenD{-3}{\left[\begin{array}{rr}{-4} &-2 \\ 7& 1 \end{array}\right]}\\\\\\ &={\left[\begin{array}{ll}{\greenD{-3} \cdot(-4)} &\greenD{-3}\cdot (-2) \\ \greenD{-3}\cdot 7& \greenD{-3}\cdot 1 \end{array}\right]}\\\\\\ &={\left[\begin{array}{rr}{12} &6 \\ -21& -3 \end{array}\right]}\end{aligned}
2) Given $C={\left[\begin{array}{rr}{-42} \\ 27 \\-3 \end{array}\right]}$, find start fraction, 1, divided by, 3, end fraction, C.
start fraction, 1, divided by, 3, end fraction, C, equals

To find start fraction, 1, divided by, 3, end fraction, C, multiply each matrix entry by start fraction, 1, divided by, 3, end fraction.
\begin{aligned}\dfrac13C&=\greenD{\dfrac{1}{3}}{\left[\begin{array}{rr}{-42} \\ 27 \\-3 \end{array}\right]}\\\\\\ &={\left[\begin{array}{l}{\greenD{\dfrac13} \cdot(-42)} \\ \greenD{\dfrac13}\cdot 27 \\\greenD{\dfrac13}\cdot (-3) \end{array}\right]}\\\\\\ &={\left[\begin{array}{rr}{-14} \\ 9 \\-1 \end{array}\right]}\end{aligned}

Recall that to add (or subtract) two matrices, we can simply add (or subtract) the corresponding entries.
For example,
\begin{aligned} &\phantom{=}{\left[\begin{array}{rr}\blueD{10} &\blueD{12} \\\blueD 6& \blueD3 \end{array}\right]}+\left[\begin{array}{rr}\goldD{1} &\goldD4 \\ \goldD{22}& \goldD7 \end{array}\right] \\\\ &={\left[\begin{array}{rr}{\blueD{10}+\goldD1} &\blueD{12}+\goldD4 \\ \blueD6+\goldD{22}& \blueD3+\goldD7 \end{array}\right]} \\\\ &=\left[\begin{array}{rr}{11} &16 \\ 28& 10 \end{array}\right]\\ \end{aligned}
Now, suppose we wanted to consider the repeated addition of a matrix.
If $A={\left[\begin{array}{rr}{4} &8 \\ 2& 1 \end{array}\right]}$, let's find A, plus, A, plus, A.
\begin{aligned} &\phantom{=}A+A+A \\\\ &= {\left[\begin{array}{rr}{4} &8 \\ 2& 1 \end{array}\right]}+{\left[\begin{array}{rr}{4} &8 \\ 2& 1 \end{array}\right]}+{\left[\begin{array}{rr}{4} &8 \\ 2& 1 \end{array}\right]}\\ \\ &={\left[\begin{array}{rr}{4+4+4} &8+8+8 \\ 2+2+2& 1+1+1 \end{array}\right]}\\\\ &={\left[\begin{array}{rr}{\greenD{3}\cdot4} &\greenD{3}\cdot8 \\ \greenD{3}\cdot2& \greenD{3}\cdot1 \end{array}\right]}\\ \\ &=\greenD{3}\cdot {\left[\begin{array}{rr}{4} &8 \\2& 1 \end{array}\right]}\\\\ &=3A \end{aligned}
Here we see that A, plus, A, plus, A, equals, 3, A.
Therefore, we can interpret scalar multiplication in the same way as we interpret multiplication with real numbers – as repeated matrix addition!

Solving matrix equations

A matrix equation is simply an equation in which the variable stands for a matrix.
For example, the equation below is a matrix equation.
$3A={\left[\begin{array}{rr}{3} &24 \\ 18& 0 \end{array}\right]}$
To solve for A, we can multiply both sides of the matrix equation by the scalar start color greenD, start fraction, 1, divided by, 3, end fraction, end color greenD.
In general, we solve a matrix equation just like we would solve any linear equation, except that the operations we perform are with matrices!

3) Solve for X.
$~~5X={\left[\begin{array}{rr}{-25} &-5 \\ 10& 20 \end{array}\right]}$
X, equals

To solve for X, we can multiply both sides of the matrix equation by the scalar start color greenD, start fraction, 1, divided by, 5, end fraction, end color greenD.
4) Solve for Y.
$~~\dfrac13Y={\left[\begin{array}{rr}{4} &-2 &~~~2 \\ 7& 1&-3 \end{array}\right]}$
Y, equals

To solve for Y, we can multiply both sides of the matrix equation by the scalar start color greenD, 3, end color greenD.

Challenge problem

5*) Solve for Z.
$~~2Z+{\left[\begin{array}{rr}{3} &6 \\ 4& 11 \end{array}\right]}={\left[\begin{array}{rr}{1} &8 \\ 10& 23 \end{array}\right]}$
Z, equals

For these more elaborate matrix equations, it is sometimes easier to make matrix substitutions.
To start, let $\greenD X=\greenD{\left[\begin{array}{rr}{3} &6 \\ 4& 11 \end{array}\right]}$ and $\blueD Y=\blueD{\left[\begin{array}{rr}{1} &8 \\ 10& 23 \end{array}\right]}$.
We can rewrite the above equation as 2, Z, plus, X, equals, Y. Solving for Z gives:
Now we can substitute matrices X and Y into the equation and solve for Z.
\begin{aligned}Z&=\dfrac12(\blueD Y-\greenD X)\\ \\ &=\dfrac12\left(\blueD{\left[\begin{array}{rr}{1} &8 \\ 10& 23 \end{array}\right]}-\greenD{\left[\begin{array}{rr}{3} &6 \\ 4& 11 \end{array}\right]}\right)\\\\\\ &=\dfrac12\left({\left[\begin{array}{rr}{1-3} &8-6 \\ 10-4& 23-11 \end{array}\right]}\right)\\\\\\ &={\dfrac12}\left({\left[\begin{array}{rr}{-2} &2 \\ 6& 12 \end{array}\right]}\right)\\\\\\ &={\left[\begin{array}{ll}{{\dfrac12}\cdot (-2)} &{\dfrac12}\cdot2 \\ {\dfrac12}\cdot 6& {\dfrac12}\cdot 12 \end{array}\right]}\\\\\\ &={\left[\begin{array}{rr}{-1} &1 \\ 3&6 \end{array}\right]} \end{aligned}
This can also be done without substituting. To do this, subtract ${\left[\begin{array}{rr}{3} &6 \\ 4& 11 \end{array}\right]}$ from both sides. Then, multiply both sides by start fraction, 1, divided by, 2, end fraction.