## Finding Scalar Multiples of a Matrix

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication involves multiplying each entry in a matrix by a scalar. A scalar multiple is any entry of a matrix that results from scalar multiplication.

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school’s current inventory is displayed in the table below.

Lab A Lab B
Computers 15 27
Computer Tables 16 34
Chairs 16 34

Converting the data to a matrix, we have

${C}_{2013}=\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right]$

To calculate how much computer equipment will be needed, we multiply all entries in matrix $C$ by 0.15.

$\left(0.15\right){C}_{2013}=\left[\begin{array}{c}\left(0.15\right)15\\ \left(0.15\right)16\\ \left(0.15\right)16\end{array}\begin{array}{c}\left(0.15\right)27\\ \left(0.15\right)34\\ \left(0.15\right)34\end{array}\right]=\left[\begin{array}{c}2.25\\ 2.4\\ 2.4\end{array}\begin{array}{c}4.05\\ 5.1\\ 5.1\end{array}\right]$

We must round up to the next integer, so the amount of new equipment needed is

$\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right]$

Adding the two matrices as shown below, we see the new inventory amounts.

$\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right]+\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right]=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right]$

This means

${C}_{2014}=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right]$

Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.

### A General Note: Scalar Multiplication

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

$A=\left[\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\ {a}_{21}& & & {a}_{22}\end{array}\right]$

the scalar multiple $cA$ is

$\begin{array}{l}cA=c\left[\begin{array}{ccc}{a}_{11}& & {a}_{12}\\ {a}_{21}& & {a}_{22}\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\ c{a}_{21}& & c{a}_{22}\end{array}\right]\hfill \end{array}$

Scalar multiplication is distributive. For the matrices $A,B$, and $C$ with scalars $a$ and $b$,

$\begin{array}{l}\\ \begin{array}{c}a\left(A+B\right)=aA+aB\\ \left(a+b\right)A=aA+bA\end{array}\end{array}$

### Example 6: Multiplying the Matrix by a Scalar

Multiply matrix $A$ by the scalar 3.

$A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right]$

### Solution

Multiply each entry in $A$ by the scalar 3.

$\begin{array}{l}3A=3\left[\begin{array}{rr}\hfill 8& \hfill 1\\ \hfill 5& \hfill 4\end{array}\right]\hfill \\ = \left[\begin{array}{rr}\hfill 3\cdot 8& \hfill 3\cdot 1\\ \hfill 3\cdot 5& \hfill 3\cdot 4\end{array}\right]\hfill \\ = \left[\begin{array}{rr}\hfill 24& \hfill 3\\ \hfill 15& \hfill 12\end{array}\right]\hfill \end{array}$

### Try It 2

Given matrix $B,\text{}$ find $-2B$ where

$B=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right]$

### Example 7: Finding the Sum of Scalar Multiples

Find the sum $3A+2B$.

$A=\left[\begin{array}{rrr}\hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill -1& \hfill 2\\ \hfill 4& \hfill 3& \hfill -6\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 1\\ \hfill 0& \hfill -3& \hfill 2\\ \hfill 0& \hfill 1& \hfill -4\end{array}\right]$

### Solution

First, find $3A,\text{}$ then $2B$.

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 3A=\left[\begin{array}{lll}3\cdot 1\hfill & 3\left(-2\right)\hfill & 3\cdot 0\hfill \\ 3\cdot 0\hfill & 3\left(-1\right)\hfill & 3\cdot 2\hfill \\ 3\cdot 4\hfill & 3\cdot 3\hfill & 3\left(-6\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{rrr}\hfill 3& \hfill -6& \hfill 0\\ \hfill 0& \hfill -3& \hfill 6\\ \hfill 12& \hfill 9& \hfill -18\end{array}\right]\hfill \end{array}$
$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 2B=\left[\begin{array}{lll}2\left(-1\right)\hfill & 2\cdot 2\hfill & 2\cdot 1\hfill \\ 2\cdot 0\hfill & 2\left(-3\right)\hfill & 2\cdot 2\hfill \\ 2\cdot 0\hfill & 2\cdot 1\hfill & 2\left(-4\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ =\left[\begin{array}{rrr}\hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill -6& \hfill 4\\ \hfill 0& \hfill 2& \hfill -8\end{array}\right]\hfill \end{array}$

Now, add $3A+2B$.

$\begin{array}{l}\hfill \\ \hfill \\ 3A+2B=\left[\begin{array}{rrr}\hfill 3& \hfill -6& \hfill 0\\ \hfill 0& \hfill -3& \hfill 6\\ \hfill 12& \hfill 9& \hfill -18\end{array}\right]+\left[\begin{array}{rrr}\hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill -6& \hfill 4\\ \hfill 0& \hfill 2& \hfill -8\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 3 - 2& \hfill -6+4& \hfill 0+2\\ \hfill 0+0& \hfill -3 - 6& \hfill 6+4\\ \hfill 12+0& \hfill 9+2& \hfill -18 - 8\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 1& \hfill -2& \hfill 2\\ \hfill 0& \hfill -9& \hfill 10\\ \hfill 12& \hfill 11& \hfill -26\end{array}\right]\hfill \end{array}$