College Algebra, Chapter 7, 7.4, Section 7.4, Problem 40

Solve the system $\left\{ \begin{array}{ccccc}
5x & -3y & +z & = & 6 \\
& 4y & -6z & = & 22 \\
7x & +10 y & & = & -13
\end{array} \right.$ using Cramer's Rule.

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
5 & -3 & 1 \\
0 & 4 & -6 \\
7 & 10 & 0
\end{array} \right| = 5 \left[ 4 \cdot 0 - (-6) \cdot 10 \right] - (-3) \left[ 0 \cdot 0 - (-6) \cdot 7 \right] + 1 (0 \cdot 10 - 4 \cdot 7) = 398
\\
\\
|D_{x}| =& \left| \begin{array}{ccc}
6 & -3 & 1 \\
22 & 4 & -6 \\
-13 & 10 & 0
\end{array} \right| = 6 \left[ 4 \cdot 0 - (-6) \cdot 10 \right] - (-3) \left[ (-6) \cdot (-13) - 22 \cdot 0 \right] + 1 \left[ 22 \cdot 10 - 4 \cdot (-13) \right] =398
\\
\\
|D_{y}| =& \left| \begin{array}{ccc}
5 & 6 & 1 \\
0 & 22 & -6 \\
7 & -13 & 0
\end{array} \right| = 5 \left[ 22 \cdot 0 - (-6) \cdot (-13) \right] - 6 \left[ (-6) \cdot 7 - 0 \cdot 0 \right] + 1 \left[ 0 \cdot (-13) - 22 \cdot 7 \right]= -796
\\
\\
|D_z| =& \left| \begin{array}{ccc}
5 & -3 & 6 \\
0 & 4 & 22 \\
7 & 10 & -13
\end{array} \right| = 5 \left[ 4 \cdot (-13) - 22 \cdot 10 \right] - (-3) \left[ 0 \cdot (-13) - 22 \cdot 7 \right] + 6 \left( 0 \cdot 10 - 4 \cdot 7 \right] = -1190

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{398}{398} = 1
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{-796}{398} = -2
\\
\\
z =& \frac{|D_z|}{|D|} = \frac{-1990}{398} = -5
\end{aligned}
\end{equation}
$