College Algebra, Chapter 7, 7.4, Section 7.4, Problem 44

Solve the system $\left\{ \begin{array}{ccccc}
2x & -y & & = & 5 \\
5x & & +3z & = & 19 \\
& 4y & +7z & = & 17
\end{array} \right.$ using Cramer's Rule.

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
2 & -1 & 0 \\
5 & 0 & 3 \\
0 & 4 & 7

\end{array} \right| = 2 \left| \begin{array}{cc}
0 & 3 \\
4 & 7
\end{array} \right| - (-1) \left| \begin{array}{cc}
5 & 3 \\
0 & 7
\end{array} \right| + 0 \left| \begin{array}{cc}
5 & 0 \\
0 & 4
\end{array} \right|
\\
\\
=& 2 (0 \cdot 7 - 3 \cdot 4) + (5 \cdot 7 - 3 \cdot 0) = 11
\\
\\
|D_x| =& \left| \begin{array}{ccc}
5 & -1 & 0 \\
19 & 0 & 3 \\
17 & 4 & 7

\end{array} \right| = 5 \left| \begin{array}{cc}
0 & 3 \\
4 & 7
\end{array} \right| - (-1) \left| \begin{array}{cc}
19 & 3 \\
17 & 7
\end{array} \right| + 0 \left| \begin{array}{cc}
19 & 0 \\
17 & 4
\end{array} \right|
\\
\\
=& 5 (0 \cdot 7 - 3 \cdot 4) + (19 \cdot 7 - 3 \cdot 17) = 22
\\
\\
|D_y| =& \left| \begin{array}{ccc}
2 & 5 & 0 \\
5 & 19 & 3 \\
0 & 17 & 7
\end{array} \right| = 2 \left| \begin{array}{cc}
19 & 3 \\
17 & 7
\end{array} \right| -5 \left| \begin{array}{cc}
5 & 3 \\
0 & 7
\end{array} \right| + 0 \left| \begin{array}{cc}
5 & 19 \\
0 & 17
\end{array} \right|
\\
\\
=& 2 (19 \cdot 7 - 3 \cdot 17) - 5 (5 \cdot 7 - 3 \cdot 0) = -11
\\
\\
|D_z| =& \left| \begin{array}{ccc}
2 & -1 & 5 \\
5 & 0 & 19 \\
0 & 4 & 17
\end{array} \right| = 2 \left| \begin{array}{cc}
0 & 19 \\
4 & 17
\end{array} \right| - (-1) \left| \begin{array}{cc}
5 & 19 \\
0 & 17
\end{array} \right| + 5 \left| \begin{array}{cc}
5 & 0 \\
0 & 4
\end{array} \right|
\\
\\
=& 2 (0 \cdot 17 - 19 \cdot 4) + (5 \cdot 17 - 19 \cdot 0) + 5 (5 \cdot 4 - 0 \cdot 0) =33

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

x =& \frac{|D_x|}{|D|} = \frac{22}{11} = 2
\\
\\
y =& \frac{|D_y|}{|D|} = \frac{-11}{11} = -1
\\
\\
z =& \frac{|D_z|}{|D|} = \frac{33}{11} = 3
\end{aligned}
\end{equation}
$