College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 54

Express the system of linear equations $
\left\{
\begin{array}{ccccc}
2x & & +3z & = & 5 \\
x & +y & +6z & = & 0 \\
3x & -y & +z & = & 5
\end{array}
\right.$ as a matrix equation then solve the matrix equation by multiplying each side by the inverse of the coefficient matrix.

The equivalent matrix equation of the system is

$\left[
\begin{array}{ccc}
2 & 0 & 3 \\
1 & 1 & 6 \\
3 & -1 & 1
\end{array}
\right] \left[ \begin{array}{c}
x \\
y \\
z
\end{array} \right] = \left[ \begin{array}{c}
5 \\
0 \\
5
\end{array} \right]$

If we let

$\displaystyle A = \left[ \begin{array}{cc}
6 & -5 \\
8 & -7
\end{array} \right] \qquad X = \left[ \begin{array}{c}
x \\
y
\end{array} \right] \qquad B = \left[ \begin{array}{c}
1 \\
-1
\end{array} \right]$

Then the matrix can be written as

$AX = B$

We solve this matrix equation by multiplying each side by the inverse of $A$


$
\begin{equation}
\begin{aligned}

AX =& B
&&
\\
A^{-1} (AX) =& A^{-1} B
&& \text{Multiply each side by } A^{-1}
\\
(A^{-1} A) X =& A^{-1} B
&& \text{Associate Property}
\\
I_3 X =& A^{-1} B
&& \text{Property of Inverses}
\\
X =& A^{-1} B
&& \text{Property of Identity Matrix}

\end{aligned}
\end{equation}
$


Solving for the inverse of $A$

Add identity matrix to the right of the matrix

$\displaystyle \left[ \begin{array}{ccc|ccc}
2 & 0 & 3 & 1 & 0 & 0 \\
1 & 1 & 6 & 0 & 1 & 0 \\
3 & -1 & 1 & 0 & 0 & 1
\end{array} \right]$

Using Gauss-Jordan Elimination

$\displaystyle \frac{1}{2} R_1$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & \displaystyle \frac{3}{2} & \displaystyle \frac{1}{2} & 0 & 0 \\
1 & 1 & 6 & 0 & 1 & 0 \\
3 & -1 & 1 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_2 - R_1 \to R_2$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & \displaystyle \frac{3}{2} & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & \displaystyle \frac{9}{2} & \displaystyle \frac{-1}{2} & 1 & 0 \\
3 & -1 & 1 & 0 & 0 & 1
\end{array} \right]$

$\displaystyle R_3 - 3 R_1 \to R_3$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & \displaystyle \frac{3}{2} & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & \displaystyle \frac{9}{2} & \displaystyle \frac{-1}{2} & 1 & 0 \\
0 & -1 & \displaystyle \frac{-7}{2} & \displaystyle \frac{-3}{2} & 0 & 1
\end{array} \right]$

$\displaystyle R_3 + R_2 \to R_3$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & \displaystyle \frac{3}{2} & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & \displaystyle \frac{9}{2} & \displaystyle \frac{-1}{2} & 1 & 0 \\
0 & 0 & 1 & -2 & 1 & 1
\end{array} \right]$

$\displaystyle R_2 - \frac{9}{2} R_3 \to R_2$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & \displaystyle \frac{3}{2} & \displaystyle \frac{1}{2} & 0 & 0 \\
0 & 1 & 0 & \displaystyle \frac{17}{2} & \displaystyle \frac{-7}{2} & \displaystyle \frac{-9}{2} \\
0 & 0 & 1 & -2 & 1 & 1
\end{array} \right]$

$\displaystyle R_1 - \frac{3}{2} R_3 \to R_1$

$\displaystyle \left[ \begin{array}{ccc|ccc}
1 & 0 & 0 & \displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} & \displaystyle \frac{-3}{2} \\
0 & 1 & 0 & \displaystyle \frac{17}{2} & \displaystyle \frac{-7}{2} & \displaystyle \frac{-9}{2} \\
0 & 0 & 1 & -2 & 1 & 1
\end{array} \right]$

So the inverse is

$\displaystyle A^{-1} = \left[ \begin{array}{ccc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} & \displaystyle \frac{-3}{2} \\
\displaystyle \frac{17}{2} & \displaystyle \frac{-7}{2} & \displaystyle \frac{-9}{2} \\
-2 & 1 & 1
\end{array} \right]$

Then,


$
\begin{equation}
\begin{aligned}

X =& A^{-1} B
\\
\\
X =& \left[ \begin{array}{ccc}
\displaystyle \frac{7}{2} & \displaystyle \frac{-3}{2} & \displaystyle \frac{-3}{2} \\
\displaystyle \frac{17}{2} & \displaystyle \frac{-7}{2} & \displaystyle \frac{-9}{2} \\
-2 & 1 & 1
\end{array} \right] \left[ \begin{array}{c}
5 \\
0 \\
5
\end{array} \right]
\\
\\
X =& \left[ \begin{array}{c}
\displaystyle \frac{7}{2} \cdot 5 + \left( \frac{-3}{2} \right) \cdot 0 + \left( \frac{-3}{2} \right) \cdot 5 \\
\displaystyle \frac{17}{2} \cdot 5 + \left( \frac{-7}{2} \right) \cdot 0 + \left( \frac{-9}{2} \right) \cdot 5 \\
-2 \cdot 5 + 1 \cdot 0 + 1 \cdot 5
\end{array} \right]
\\
\\
X =& \left[ \begin{array}{c}
10 \\
20 \\
-5
\end{array} \right]


\end{aligned}
\end{equation}
$