# 7.5 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.
$\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
hi vedant can u help me with some assignments
Solomon
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
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Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
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