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} = ⎡ ⎢ ⎣ \begin{matrix} 151616 \end{matrix} \begin{matrix} 273434 \end{matrix} ⎤ ⎥ ⎦

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

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

(0.15) C_{2013} = ⎡ ⎢ ⎣ \begin{matrix} (0.15) 15 (0.15) 16 (0.15) 16 \end{matrix} \begin{matrix} (0.15) 27 (0.15) 34 (0.15) 34 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 2.25 2.4 2.4 \end{matrix} \begin{matrix} 4.05 5.1 5.1 \end{matrix} ⎤ ⎥ ⎦

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

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

⎡ ⎢ ⎣ \begin{matrix} 333 \end{matrix} \begin{matrix} 566 \end{matrix} ⎤ ⎥ ⎦

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

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

⎡ ⎢ ⎣ \begin{matrix} 151616 \end{matrix} \begin{matrix} 273434 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} 333 \end{matrix} \begin{matrix} 566 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 181919 \end{matrix} \begin{matrix} 324040 \end{matrix} ⎤ ⎥ ⎦

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

This means

C_{2014} = ⎡ ⎢ ⎣ \begin{matrix} 181919 \end{matrix} \begin{matrix} 324040 \end{matrix} ⎤ ⎥ ⎦

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

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 = [\begin{matrix} a_{11} & a_{12} a_{21} & a_{22} \end{matrix}]

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

the scalar multiple $cA$ is

\begin{matrix} c A = c [\begin{matrix} a_{11} & a_{12} a_{21} & a_{22} \end{matrix}] = [\begin{matrix} c a_{11} & c a_{12} c a_{21} & c a_{22} \end{matrix}] \end{matrix}

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

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

\begin{matrix} \begin{matrix} a (A + B) = a A + a B (a + b) A = a A + b A \end{matrix} \end{matrix}

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

Example 6: Multiplying the Matrix by a Scalar

Multiply matrix $A$ by the scalar 3.

A = [\begin{matrix} 8 & 1 5 & 4 \end{matrix}]

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

Solution

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

\begin{matrix} 3 A = 3 [\begin{matrix} 8 & 1 5 & 4 \end{matrix}] = [\begin{matrix} 3 \cdot 8 & 3 \cdot 1 3 \cdot 5 & 3 \cdot 4 \end{matrix}] = [\begin{matrix} 24 & 3 15 & 12 \end{matrix}] \end{matrix}

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

Try It 2

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

B = [\begin{matrix} 4 & 1 3 & 2 \end{matrix}]

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

Solution

Example 7: Finding the Sum of Scalar Multiples

Find the sum $3A+2B$ .

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 0 0 & - 1 & 2 4 & 3 & - 6 \end{matrix} ⎤ ⎥ ⎦ and B = ⎡ ⎢ ⎣ \begin{matrix} - 1 & 2 & 1 0 & - 3 & 2 0 & 1 & - 4 \end{matrix} ⎤ ⎥ ⎦

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

Solution

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

\begin{matrix} \begin{matrix} 3 A = ⎡ ⎢ ⎣ \begin{matrix} 3 \cdot 1 & 3 (- 2) & 3 \cdot 0 3 \cdot 0 & 3 (- 1) & 3 \cdot 2 3 \cdot 4 & 3 \cdot 3 & 3 (- 6) \end{matrix} ⎤ ⎥ ⎦ \end{matrix} = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 6 & 0 0 & - 3 & 6 12 & 9 & - 18 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}

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

\begin{matrix} \begin{matrix} 2 B = ⎡ ⎢ ⎣ \begin{matrix} 2 (- 1) & 2 \cdot 2 & 2 \cdot 1 2 \cdot 0 & 2 (- 3) & 2 \cdot 2 2 \cdot 0 & 2 \cdot 1 & 2 (- 4) \end{matrix} ⎤ ⎥ ⎦ \end{matrix} = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 4 & 2 0 & - 6 & 4 0 & 2 & - 8 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}

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

Now, add $3A+2B$ .

\begin{matrix} 3 A + 2 B = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 6 & 0 0 & - 3 & 6 12 & 9 & - 18 \end{matrix} ⎤ ⎥ ⎦ + ⎡ ⎢ ⎣ \begin{matrix} - 2 & 4 & 2 0 & - 6 & 4 0 & 2 & - 8 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 3 - 2 & - 6 + 4 & 0 + 2 0 + 0 & - 3 - 6 & 6 + 4 12 + 0 & 9 + 2 & - 18 - 8 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 2 & 2 0 & - 9 & 10 12 & 11 & - 26 \end{matrix} ⎤ ⎥ ⎦ \end{matrix}

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

Matrices and Matrix Operations