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# Multiplying Matrices by a Scalar

## Multiply and distribute single values across matrices

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Multiplying Matrices by a Scalar

A group of 8 kids, a group of 8 adults, and a group of 8 seniors are attending a movie.

For a matinee movie, the movie theater charges the following prices:

Kids: $5 Adults:$8 Seniors: $6 For the same movie at night, the theater charges the following prices: Kids:$7 Adults: $10 Seniors:$8

How could we determine the total amount each group will be charged for each type of movie?

### Guidance

A matrix can be multiplied by a scalar. A scalar is a real number in matrix algebra-not a matrix. To multiply a matrix by a scalar, each element in the matrix is multiplied by the scalar as shown below:

$\ k \begin{bmatrix}a & b\\c & d \end{bmatrix} = \begin{bmatrix}ka & kb\\kc & kd \end{bmatrix}, \ \text{where} \ k \ \text{is a scalar}.$

#### Investigation: Distributive Property of Scalar Multiplication

Let’s investigate what happens if we distribute the multiplication of a scalar of the addition of two matrices. Consider the matrix expression: $3 \left( \begin{bmatrix}2\\-5 \end{bmatrix} + \begin{bmatrix}-3\\6 \end{bmatrix} \right)$

Method 1: Perform the addition inside the parenthesis first and then multiply by the scalar:

$3\left( \begin{bmatrix}2\\-5 \end{bmatrix} + \begin{bmatrix}-3\\6 \end{bmatrix} \right) = 3\begin{bmatrix}-1\\1 \end{bmatrix} = \begin{bmatrix}-3\\3 \end{bmatrix}$

Method 2: Distribute the scalar into both matrices and then add:

$3 \left( \begin{bmatrix}2\\-5 \end{bmatrix} + \begin{bmatrix}-3\\6 \end{bmatrix} \right) = \begin{bmatrix}6\\-15 \end{bmatrix} + \begin{bmatrix}-9\\18 \end{bmatrix} = \begin{bmatrix}-3\\3 \end{bmatrix}$

The results are equivalent. We can conjecture that the Distributive Property of Multiplication over Addition is true for scalar multiplication of matrices. This property can be extended to include distribution of scalar multiplication over subtraction as well.

#### Properties of Scalar Multiplication

Distributive Property of Addition: $k(A+B) = kA+kB$

Distributive Property of Subtraction: $k(A-B) = kA-KB$

#### Example A

Perform the scalar multiplication: $2\begin{bmatrix}-4 & \frac{1}{2}\\-1 & 3 \end{bmatrix}$

Solution: In this case we just need to multiply each element of the matrix by 2.

$\begin{bmatrix}2(-4) & 2\left( \frac{1}{2} \right)\\2(-1) & 2(3) \end{bmatrix} = \begin{bmatrix}-8 & 1\\-2 & 6 \end{bmatrix}$

#### Example B

Perform the indicated operation: $3\left( \begin{bmatrix}2\\-1 \end{bmatrix} + \begin{bmatrix}\frac{2}{3}\\4 \end{bmatrix} \right)$

Solution: This time we need to decide whether to distribute the 3 inside the parenthesis or add first and then multiply by 3. It is possible to complete this problem either way. However, careful observation allows us to observe that there is a fraction inside the second matrix. By distributing the 3 first, we can eliminate this fraction and make the addition easier as shown below.

$\left( 3\begin{bmatrix}2\\-1 \end{bmatrix} + 3\begin{bmatrix}\frac{2}{3}\\4 \end{bmatrix} \right) = \left( \begin{bmatrix}6\\-3 \end{bmatrix} + \begin{bmatrix}2\\12 \end{bmatrix} \right) = \begin{bmatrix}8\\9 \end{bmatrix}$

#### Example C

Perform the indicated operation: $\frac{1}{2} \left( \begin{bmatrix}7 & -1\\2 & 8 \end{bmatrix} + \begin{bmatrix}-3 & 5\\2 & 0 \end{bmatrix} \right)$

Solution: Again, we need to decide whether to do the multiplication or addition first. Here, it turns out to be easier to add first and then multiply as shown below.

$\frac{1}{2} \begin{bmatrix}7+-3 & -1+5\\2+2 & 8+0 \end{bmatrix} = \frac{1}{2}\begin{bmatrix}4 & 4\\4 & 8 \end{bmatrix} = \begin{bmatrix}2 & 2\\2 & 4 \end{bmatrix}$

Not that this problem could have been solved in the other order, but we would have had to deal with fractions.

Intro Problem Revisit We could organize the data in a matrix and multiply it by the scalar 8.

$& \quad K \quad A \quad S\\\begin{matrix}\text{Matinee \ }\\\text{ \quad Night \ }\end{matrix} & 8\begin{bmatrix}5 & 8 & 6\\7 & 10 & 8\end{bmatrix} =$ $& \quad K \quad A \quad S\\\begin{matrix}\text{Matinee \ }\\\text{ \quad Night \ }\end{matrix} & \begin{bmatrix}40 & 64 & 48\\56 & 80 & 64\end{bmatrix}$

We can now easily see that the group of adults will be charged $64 for a matinee, the group of kids will be charged$56 at night, etc.

### Guided Practice

Perform the indicated operations in the following problems.

1. $\frac{2}{3} \begin{bmatrix}0\\6\\9 \end{bmatrix}$

2. $-\frac{2}{3} \left( \begin{bmatrix}2 & -3 & 5 \end{bmatrix} - \begin{bmatrix}-1 & 0 & 2 \end{bmatrix} \right)$

3. $12\left( \begin{bmatrix}\frac{3}{4} & -\frac{2}{3}\\\frac{1}{6} & 2 \end{bmatrix} + \begin{bmatrix}1 & \frac{5}{6}\\\frac{2}{3} & \frac{5}{4} \end{bmatrix} \right)$

1. Multiply each element inside the matrix by $\frac{2}{3}$ :

$\begin{bmatrix}\frac{2}{3}(0)\\\frac{2}{3}(6)\\\frac{2}{3}(9) \end{bmatrix} = \begin{bmatrix}0\\4\\6 \end{bmatrix}$

2. If we subtract what is inside the parenthesis first, we can avoid fractions:

$- \frac{2}{3} \left( \begin{bmatrix}2-(-1) & -3-0 & 5-2 \end{bmatrix} \right) = - \frac{2}{3} \begin{bmatrix}3 & -3 & 3 \end{bmatrix} = \begin{bmatrix}-2 & 2 & -2 \end{bmatrix}$

3. If we distribute first this time, we can avoid fractions:

$\left( 12\begin{bmatrix}\frac{3}{4} & -\frac{2}{3}\\\frac{1}{6} & 2 \end{bmatrix} + 12\begin{bmatrix}1 & \frac{5}{6}\\\frac{2}{3} & \frac{5}{4} \end{bmatrix} \right) = \begin{bmatrix}9 & -8\\2 & 24 \end{bmatrix} + \begin{bmatrix}12 & 10\\8 & 15 \end{bmatrix} = \begin{bmatrix}21 & 2\\10 & 39 \end{bmatrix}$

### Practice

Perform the indicated operations, if possible.

1. .
$3\begin{bmatrix}2 & -1\\8 & 0 \end{bmatrix}$
1. .
$-2\begin{bmatrix}-6 & 8\\5 & -2 \end{bmatrix}$
1. .
$\frac{2}{3}\begin{bmatrix}12\\6 \end{bmatrix}$
1. .
$-\frac{3}{2}\begin{bmatrix}8 & 0 & -4\\-6 & 2 & 10 \end{bmatrix}$
1. .
$5\begin{bmatrix}-3 & 1 & 2 \end{bmatrix}$
1. .
$-1 \begin{bmatrix}2 & -3\\5 & 1\\-8 & -10 \end{bmatrix}$
1. .
$2 \begin{bmatrix}1 & 2\\4 & -1\end{bmatrix} + \begin{bmatrix}4 & -2\\-3 & -7 \end{bmatrix}$
1. .
$3 \begin{bmatrix}4 & -5 & -1 \end{bmatrix} + 4\begin{bmatrix}8 & -1 & 5\end{bmatrix}$
1. .
$-2 \begin{bmatrix}2 & -3 & 0 \\-1 & -4 & 3 \\-1 & -1 & 4 \end{bmatrix} - (-1)\begin{bmatrix}3 & 3 & -2\\-4 & -10 & 8\end{bmatrix}$
1. .
$-2 \left( \begin{bmatrix}\frac{1}{2} & \frac{5}{2} \end{bmatrix} - \begin{bmatrix}3 & -1 \end{bmatrix} \right)$
1. .
$-\frac{1}{3} \left( \begin{bmatrix}5\\-2 \end{bmatrix} - \begin{bmatrix}2\\4 \end{bmatrix} \right)$
1. .
$3\begin{bmatrix}8 & -2\\\frac{1}{3} & 5 \end{bmatrix} + \begin{bmatrix}2 & 4\\-6 & -1 \end{bmatrix}$
1. .
$2\begin{bmatrix}\frac{3}{7}\\\frac{3}{4} \end{bmatrix}$
1. .
$-6\left( \begin{bmatrix}\frac{1}{3} & 0\\2 & - \frac{2}{3} \end{bmatrix} + \begin{bmatrix}5 & 1\\- \frac{1}{6} & 2 \end{bmatrix} \right)$
1. .
$\begin{bmatrix}-2\\3\\8 \end{bmatrix} -5\left( \begin{bmatrix}\frac{1}{5}\\- \frac{2}{5}\\\frac{11}{5} \end{bmatrix} + \begin{bmatrix}-2\\\frac{1}{5}\\- \frac{3}{5} \end{bmatrix} \right)$