College Algebra, Chapter 7, 7.4, Section 7.4, Problem 42

Solve the system $\left\{ \begin{array}{ccccc}
-2a & & + c & = & 2 \\
a & + 2b & -c & = & 9 \\
3a & + 5b & +2c & = & 22
\end{array} \right.$ using Cramer's Rule.

For this system we have


$
\begin{equation}
\begin{aligned}

|D| =& \left| \begin{array}{ccc}
-2 & 0 & 1 \\
1 & 2 & -1 \\
3 & 5 & 2
\end{array} \right| = -2 \left[ 2 \cdot 2 - (-1) \cdot 5 \right] + 1 \left( 1 \cdot 5 - 2 \cdot 3 \right)
= -19
\\
\\
|D_a| =& \left| \begin{array}{ccc}
2 & 0 & 1 \\
9 & 2 & -1 \\
22 & 5 & 2
\end{array} \right| = 2 \left[ 2 \cdot 2 - (-1) \cdot 5 \right] + 1 \left( 9 \cdot 5 - 2 \cdot 22 \right)
= 19
\\
\\
|D_b| =& \left| \begin{array}{ccc}
-2 & 2 & 1 \\
1 & 9 & -1 \\
3 & 22 & 2
\end{array} \right| = -2 \left[ 9 \cdot 2 - (-1) \cdot 22 \right] - 2 \left[ 1 \cdot 2 - (-1) \cdot 3 \right] + 1 \left( 1 \cdot 22 - 9 \cdot 3 \right)
= -95
\\
\\
|D_c| =& \left| \begin{array}{ccc}
-2 & 0 & 2 \\
1 & 2 & 9 \\
3 & 5 & 22
\end{array} \right| = -2 \left( 2 \cdot 22 - 9 \cdot 5 \right) + 2 \left( 1 \cdot 5 - 2 \cdot 3 \right)
= 0

\end{aligned}
\end{equation}
$


The solution is


$
\begin{equation}
\begin{aligned}

a =& \frac{|D_a|}{|D|} = \frac{19}{-19} = -1
\\
\\
b =& \frac{|D_b|}{|D|} = \frac{-95}{-19} = 5
\\
\\
c =& \frac{|D_c|}{|D|} = \frac{0}{-19} = 0
\end{aligned}
\end{equation}
$

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