# 9.5 Matrices and matrix operations  (Page 3/10)

 Page 3 / 10
Lab A Lab B
Computers 15 27
Computer Tables 16 34
Chairs 16 34

Converting the data to a matrix, we have

${C}_{2013}=\left[\begin{array}{c}15\\ 16\\ 16\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}27\\ 34\\ 34\end{array}\right]$

To calculate how much computer equipment will be needed, we multiply all entries in matrix $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ by 0.15.

$\left(0.15\right){C}_{2013}=\left[\begin{array}{c}\left(0.15\right)15\\ \left(0.15\right)16\\ \left(0.15\right)16\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\left(0.15\right)27\\ \left(0.15\right)34\\ \left(0.15\right)34\end{array}\right]=\left[\begin{array}{c}2.25\\ 2.4\\ 2.4\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}4.05\\ 5.1\\ 5.1\end{array}\right]$

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

$\left[\begin{array}{c}3\\ 3\\ 3\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}5\\ 6\\ 6\end{array}\right]$

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

$\left[\begin{array}{c}15\\ 16\\ 16\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}27\\ 34\\ 34\end{array}\right]+\left[\begin{array}{c}3\\ 3\\ 3\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}5\\ 6\\ 6\end{array}\right]=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}32\\ 40\\ 40\end{array}\right]$

This means

${C}_{2014}=\left[\begin{array}{c}18\\ 19\\ 19\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}32\\ 40\\ 40\end{array}\right]$

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.

## Scalar multiplication

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given

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

the scalar multiple $\text{\hspace{0.17em}}cA\text{\hspace{0.17em}}$ is

Scalar multiplication is distributive. For the matrices $\text{\hspace{0.17em}}A,B,$ and $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ with scalars $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b,$

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

## Multiplying the matrix by a scalar

Multiply matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the scalar 3.

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

Multiply each entry in $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the scalar 3.

Given matrix $\text{\hspace{0.17em}}B,\text{}$ find $\text{\hspace{0.17em}}-2B\text{\hspace{0.17em}}$ where

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

$-2B=\left[\begin{array}{cc}-8& -2\\ -6& -4\end{array}\right]$

## Finding the sum of scalar multiples

Find the sum $\text{\hspace{0.17em}}3A+2B.$

First, find $\text{\hspace{0.17em}}3A,\text{}$ then $\text{\hspace{0.17em}}2B.$

$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 3A=\left[\begin{array}{lll}3\cdot 1\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\left(-2\right)\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\cdot 0\hfill \\ 3\cdot 0\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\left(-1\right)\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\cdot 2\hfill \\ 3\cdot 4\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\cdot 3\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\left(-6\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{rrr}\hfill 3& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}-6& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \hfill 0& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}-3& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}6\\ \hfill 12& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}9& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}-18\end{array}\right]\hfill \end{array}$
$\begin{array}{l}\begin{array}{l}\hfill \\ \hfill \\ 2B=\left[\begin{array}{lll}2\left(-1\right)\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\cdot 2\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\cdot 1\hfill \\ 2\cdot 0\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(-3\right)\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\cdot 2\hfill \\ 2\cdot 0\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\cdot 1\hfill & \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\left(-4\right)\hfill \end{array}\right]\hfill \end{array}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{rrr}\hfill -2& \hfill 4& \hfill 2\\ \hfill 0& \hfill -6& \hfill 4\\ \hfill 0& \hfill 2& \hfill -8\end{array}\right]\hfill \end{array}$

Now, add $\text{\hspace{0.17em}}3A+2B.$

## Finding the product of two matrices

In addition to multiplying a matrix by a scalar, we can multiply two matrices. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. If $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is an matrix and $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ is an matrix, then the product matrix $\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$ is an matrix. For example, the product $\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$ is possible because the number of columns in $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ is the same as the number of rows in $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ If the inner dimensions do not match, the product is not defined.

We multiply entries of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ with entries of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ according to a specific pattern as outlined below. The process of matrix multiplication becomes clearer when working a problem with real numbers.

To obtain the entries in row $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}AB,\text{}$ we multiply the entries in row $\text{\hspace{0.17em}}i\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by column $\text{\hspace{0.17em}}j\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ and add. For example, given matrices $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B,\text{}$ where the dimensions of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are and the dimensions of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ are the product of $\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$ will be a matrix.

Multiply and add as follows to obtain the first entry of the product matrix $\text{\hspace{0.17em}}AB.$

1. To obtain the entry in row 1, column 1 of $\text{\hspace{0.17em}}AB,\text{}$ multiply the first row in $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the first column in $\text{\hspace{0.17em}}B,$ and add.
$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}$
2. To obtain the entry in row 1, column 2 of $\text{\hspace{0.17em}}AB,\text{}$ multiply the first row of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the second column in $\text{\hspace{0.17em}}B,$ and add.
$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}$
3. To obtain the entry in row 1, column 3 of $\text{\hspace{0.17em}}AB,\text{}$ multiply the first row of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the third column in $\text{\hspace{0.17em}}B,$ and add.
$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$

#### Questions & Answers

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