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.
Lab A | Lab B | |
---|---|---|
Computers | [latex]15[/latex] | [latex]27[/latex] |
Computer Tables | [latex]16[/latex] | [latex]34[/latex] |
Chairs | [latex]16[/latex] | [latex]34[/latex] |
Converting the data to a matrix, we have the computer inventory in fall 2013 given by
[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 in 2014, we multiply all entries in matrix [latex]C[/latex] by [latex]0.15[/latex].
[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 [latex]18[/latex] computers, [latex]19[/latex] computer tables, and [latex]19[/latex] chairs; Lab B will have [latex]32[/latex] computers, [latex]40[/latex] computer tables, and [latex]40[/latex] chairs.
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}{ll}cA & =c\left[\begin{array}{ccc}{a}_{11}& & {a}_{12}\\ {a}_{21}& & {a}_{22}\end{array}\right]\hfill \\ & =\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]
[latex]A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right][/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]
Find the sum [latex]3A+2B[/latex].