College Algebra, Chapter 7, 7.4, Section 7.4, Problem 46

Solve the system $\left\{ \begin{array}{ccccc}
2x & -5y & & = & 4 \\
x & +y & -z & = & 8 \\
3x & & +5z & = & 0
\end{array} \right.$ using Cramer's Rule.

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
2 & -5 & 0 \\
1 & 1 & -1 \\
3 & 0 & 5
\end{array} \right| = 2
\left| \begin{array}{cc}
1 & -1 \\
0 & 5
\end{array} \right| - (-5) \left| \begin{array}{cc}
1 & -1 \\
3 & 5
\end{array} \right| + 0 \left| \begin{array}{cc}
1 & 1 \\
3 & 0
\end{array} \right|
\\
\\
=& 2 [1 \cdot 5 - (-1) \cdot 0] + 5 [1 \cdot 5 - (-1) \cdot 3]
\\
\\
=& 50
\\
\\
|D_x| =& \left| \begin{array}{ccc}
4 & -5 & 0 \\
8 & 1 & -1 \\
0 & 0 & 5
\end{array} \right| = 4 \left| \begin{array}{cc}
1 & -1 \\
0 & 5
\end{array} \right| - (-5) \left| \begin{array}{cc}
8 & -1 \\
0 & 5
\end{array} \right| + 0 \left| \begin{array}{cc}
8 & 1 \\
0 & 0
\end{array} \right|
\\
\\
=& 4 [1 \cdot 5 - (-1) \cdot 0] + 5 [8 \cdot 5 - (-1) \cdot 0]
\\
\\
=& 220
\\
\\
|D_y| =& \left| \begin{array}{ccc}
2 & 4 & 0 \\
1 & 8 & -1 \\
3 & 0 & 5
\end{array} \right| = 2 \left| \begin{array}{cc}
8 & -1 \\
0 & 5
\end{array} \right| - 4 \left| \begin{array}{cc}
1 & -1 \\
3 & 5
\end{array} \right| + 0 \left| \begin{array}{cc}
1 & 8 \\
3 & 0
\end{array} \right|
\\
\\
=& 2 [8 \cdot 5 - (-1) \cdot 0] - 4 [1 \cdot 5 - (-1) \cdot 3]
\\
\\
=& 48
\\
\\
|D_z| =& \left| \begin{array}{ccc}
2 & -5 & 4 \\
1 & 1 & 8 \\
3 & 0 & 0
\end{array} \right| = 2 \left| \begin{array}{cc}
1 & 8 \\
0 & 0
\end{array} \right| - (-5) \left| \begin{array}{cc}
1 & 8 \\
3 & 0
\end{array} \right| + 4 \left| \begin{array}{cc}
1 & 1 \\
3 & 0
\end{array} \right|
\\
\\
=& 2 (1 \cdot 0 - 8 \cdot 0) + 5 (1 \cdot 0 - 8 \cdot 3) + 4 (1 \cdot 0 - 1 \cdot 3)
\\
\\
=& -132



\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{220}{50} = \frac{22}{5}
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{48}{50} = \frac{24}{25}
\\
\\
z =& \frac{|D_z|}{|D|} = \frac{-132}{50} = \frac{-66}{25}
\end{aligned}
\end{equation}
$