# 0.2 Matrices  (Page 3/11)

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Given the matrices $E$ , $F$ , $G$ and $H$ , below

$E=\left[\begin{array}{cc}1& 2\\ 4& 2\\ 3& 1\end{array}\right]F=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]G=\left[\begin{array}{cc}4& 1\end{array}\right]H=\left[\begin{array}{c}-3\\ -1\end{array}\right]$

Find, if possible.

1. $\text{EF}$
2. $\text{FE}$
3. $\text{FH}$
4. $\text{GH}$
1. To find $\text{EF}$ , we multiply the first row $\left[\begin{array}{cc}1& 2\end{array}\right]$ of $E$ with the columns $\left[\begin{array}{c}2\\ 3\end{array}\right]$ and $\left[\begin{array}{c}-1\\ 2\end{array}\right]$ of the matrix $F$ , and then repeat the process by multiplying the other two rows of $E$ with these columns of $F$ . The result is as follows:

$\begin{array}{ccc}\text{EF}& =& \left[\begin{array}{cc}1& 2\\ 4& 2\\ 3& 1\end{array}\right]\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}1\cdot 2+2\cdot 3& 1\cdot -1+2\cdot 2\\ 4\cdot 2+2\cdot 3& 4\cdot -1+2\cdot 2\\ 3\cdot 2+1\cdot 3& 3\cdot -1+1\cdot 2\end{array}\right]=\left[\begin{array}{cc}8& 3\\ \text{14}& 0\\ 9& -1\end{array}\right]\hfill \end{array}$
2. The product $\text{FE}$ is not possible because the matrix $F$ has two entries in each row, while the matrix $E$ has three entries in each column. In other words, the matrix $F$ has two columns, while the matrix $E$ has three rows.

3. $\text{FH}=\left[\begin{array}{cc}2& -1\\ 3& 2\end{array}\right]\left[\begin{array}{c}-3\\ -1\end{array}\right]=\left[\begin{array}{c}2\cdot -3+-1\cdot -1\\ 3\cdot -3+2\cdot -1\end{array}\right]=\left[\begin{array}{c}-5\\ -\text{11}\end{array}\right]$

4. $\text{GH}=\left[\begin{array}{cc}4& 1\end{array}\right]\left[\begin{array}{c}-3\\ -1\end{array}\right]=\left[4\cdot -3+1\cdot -1\right]=\left[-\text{13}\right]$

We summarize matrix multiplication as follows:

In order for product $\text{AB}$ to exist, the number of columns of $A$ , must equal the number of rows of $B$ . If matrix $A$ is of dimension $m×n$ and $B$ of dimension $n×p$ , the product will have the dimension $m×p$ . Furthermore, matrix multiplication is not commutative.

Given the matrices $R$ , $S$ , and $T$ below.

$R=\left[\begin{array}{ccc}1& 0& 2\\ 2& 1& 5\\ 2& 3& 1\end{array}\right]S=\left[\begin{array}{ccc}0& -1& 2\\ 3& 1& 0\\ 4& 2& 1\end{array}\right]T=\left[\begin{array}{ccc}-2& 3& 0\\ -3& 2& 2\\ -1& 1& 0\end{array}\right]$

Find $2\text{RS}-3\text{ST}$ .

We multiply the matrices $R$ and $S$ .

$\text{RS}=\left[\begin{array}{ccc}8& 3& 4\\ \text{23}& 9& 9\\ \text{13}& 3& 5\end{array}\right]$
$2\text{RS}=2\left[\begin{array}{ccc}8& 3& 4\\ \text{23}& 9& 9\\ \text{13}& 3& 5\end{array}\right]=\left[\begin{array}{ccc}\text{16}& 6& 8\\ \text{46}& \text{18}& \text{18}\\ \text{26}& 6& \text{10}\end{array}\right]$
$\text{ST}=\left[\begin{array}{ccc}1& 0& -2\\ -9& \text{11}& 2\\ -\text{15}& \text{17}& 4\end{array}\right]$
$3\text{ST}=3\left[\begin{array}{ccc}1& 0& -2\\ -9& \text{11}& 2\\ -\text{15}& \text{17}& 4\end{array}\right]=\left[\begin{array}{ccc}3& 0& -6\\ -\text{27}& \text{33}& 6\\ -\text{45}& \text{51}& \text{12}\end{array}\right]$
$2\text{RS}-3\text{ST}=\left[\begin{array}{ccc}\text{16}& 6& 8\\ \text{46}& \text{18}& \text{18}\\ \text{26}& 6& \text{10}\end{array}\right]-\left[\begin{array}{ccc}3& 0& -6\\ -\text{27}& \text{33}& 6\\ -\text{45}& \text{51}& \text{12}\end{array}\right]=\left[\begin{array}{ccc}\text{13}& 6& \text{14}\\ \text{73}& -\text{15}& \text{12}\\ \text{71}& -\text{45}& -2\end{array}\right]$

In this chapter, we will be using matrices to solve linear systems. In [link] , we will be asked to express linear systems as the matrix equation $\text{AX}=B$ , where $A$ , $X$ , and $B$ are matrices. The matrix $A$ is called the coefficient matrix .

Verify that the system of two linear equations with two unknowns:

$\text{ax}+\text{by}=h$
$\text{cx}+\text{dy}=k$

can be written as $\text{AX}=B$ , where

If we multiply the matrices $A$ and $X$ , we get

$\text{AX}=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}\text{ax}+\text{by}\\ \text{cx}+\text{dy}\end{array}\right]$

If $\text{AX}=B$ then

$\left[\begin{array}{c}\text{ax}+\text{by}\\ \text{cx}+\text{dy}\end{array}\right]=\left[\begin{array}{c}h\\ k\end{array}\right]$

If two matrices are equal, then their corresponding entries are equal. Therefore, it follows that

$\text{ax}+\text{by}=h$
$\text{cx}+\text{dy}=k$

Express the following system as $\text{AX}=B$ .

$2x+\text{3y}–4z=5$
$3x+4y-5z=6$
$5x-6z=7$

The above system of equations can be expressed in the form $\text{AX}=B$ as shown below.

$\left[\begin{array}{ccc}2& 3& -4\\ 3& 4& -5\\ 5& 0& -6\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$

## Systems of linear equations; gauss-jordan method

In this section, we learn to solve systems of linear equations using a process called the Gauss-Jordan method. The process begins by first expressing the system as a matrix, and then reducing it to an equivalent system by simple row operations. The process is continued until the solution is obvious from the matrix. The matrix that represents the system is called the augmented matrix , and the arithmetic manipulation that is used to move from a system to a reduced equivalent system is called a row operation .

Write the following system as an augmented matrix.

$2x+3y-4z=5$
$3x+4y-5z=-6$
$4x+5y-6z=7$

We express the above information in matrix form. Since a system is entirely determined by its coefficient matrix and by its matrix of constant terms, the augmented matrix will include only the coefficient matrix and the constant matrix. So the augmented matrix we get is as follows:

$\left[\begin{array}{ccccc}2& 3& -4& \mid & 5\\ 3& 4& -5& \mid & -6\\ 4& 5& -6& \mid & 7\end{array}\right]$

In the [link] , we expressed the system of equations as $\text{AX}=B$ , where $A$ represented the coefficient matrix, and $B$ the matrix of constant terms. As an augmented matrix, we write the matrix as $\left[A\mid B\right]$ . It is clear that all of the information is maintained in this matrix form, and only the letters $x$ , $y$ and $z$ are missing. A student may choose to write $x$ , $y$ and $z$ on top of the first three columns to help ease the transition.

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