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
[latex]{C}_{2013}=\left[\begin{array}{c}15\\ 16\\ 16\end{array}\begin{array}{c}27\\ 34\\ 34\end{array}\right][/latex]
To calculate how much computer equipment will be needed, we multiply all entries in matrix [latex]C[/latex] by 0.15.
[latex]\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][/latex]
We must round up to the next integer, so the amount of new equipment needed is
[latex]\left[\begin{array}{c}3\\ 3\\ 3\end{array}\begin{array}{c}5\\ 6\\ 6\end{array}\right][/latex]
Adding the two matrices as shown below, we see the new inventory amounts.
[latex]\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][/latex]
This means
[latex]{C}_{2014}=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\begin{array}{c}32\\ 40\\ 40\end{array}\right][/latex]
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
[latex]A=\left[\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\ {a}_{21}& & & {a}_{22}\end{array}\right][/latex]
the scalar multiple [latex]cA[/latex] is
[latex]\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}[/latex]
Scalar multiplication is distributive. For the matrices [latex]A,B[/latex], and [latex]C[/latex] with scalars [latex]a[/latex] and [latex]b[/latex],
[latex]\begin{array}{l}\\ \begin{array}{c}a\left(A+B\right)=aA+aB\\ \left(a+b\right)A=aA+bA\end{array}\end{array}[/latex]
Example 6: Multiplying the Matrix by a Scalar
Multiply matrix [latex]A[/latex] by the scalar 3.
[latex]A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right][/latex]
Solution
Multiply each entry in [latex]A[/latex] by the scalar 3.
[latex]\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}[/latex]
Try It 2
Given matrix [latex]B,\text{}[/latex] find [latex]-2B[/latex] where
[latex]B=\left[\begin{array}{cc}4& 1\\ 3& 2\end{array}\right][/latex]
Example 7: Finding the Sum of Scalar Multiples
Find the sum [latex]3A+2B[/latex].
[latex]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][/latex]
Solution
First, find [latex]3A,\text{}[/latex] then [latex]2B[/latex].
[latex]\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}[/latex]
[latex]\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}[/latex]
Now, add [latex]3A+2B[/latex].
[latex]\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}[/latex]