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Study Guides > College Algebra

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
C2013=[151616273434]{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 CC by 0.15.
(0.15)C2013=[(0.15)15(0.15)16(0.15)16(0.15)27(0.15)34(0.15)34]=[2.252.42.44.055.15.1]\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
[333566]\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.
[151616273434]+[333566]=[181919324040]\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
C2014=[181919324040]{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=[a11a12a21a22]A=\left[\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\ {a}_{21}& & & {a}_{22}\end{array}\right]
the scalar multiple cAcA is
cA=c[a11a12a21a22] =[ca11ca12ca21ca22]\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,BA,B, and CC with scalars aa and bb,
a(A+B)=aA+aB(a+b)A=aA+bA\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 AA by the scalar 3.
A=[8154]A=\left[\begin{array}{cc}8& 1\\ 5& 4\end{array}\right]

Solution

Multiply each entry in AA by the scalar 3.
3A=3[8154]=[38313534]=[2431512]\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,B,\text{} find 2B-2B where
B=[4132]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+2B3A+2B.
A=[120012436] and B=[121032014]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,3A,\text{} then 2B2B.
3A=[313(2)30303(1)3234333(6)]=[36003612918]\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}
2B=[2(1)2221202(3)2220212(4)]=[242064028]\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+2B3A+2B.
3A+2B=[36003612918]+[242064028] =[326+40+20+0366+412+09+2188] =[1220910121126]\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}

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