College Algebra, Chapter 7, 7.4, Section 7.4, Problem 28

Find the determinant of the matrix $\displaystyle A = \left| \begin{array}{cccc}
-2 & 3 & -1 & 7 \\
4 & 6 & -2 & 3 \\
7 & 7 & 0 & 5 \\
3 & -12 & 4 & 0
\end{array} \right|$, using row/column operations.

If we add 3 times column 3 to column 2, we get

$\displaystyle \left| \begin{array}{cccc}
-2 & 0 & -1 & 7 \\
4 & 0 & -2 & 3 \\
7 & 7 & 0 & 5 \\
3 & 0 & 4 & 0
\end{array} \right|$

So,

$\displaystyle \det (A) = -7 \left| \begin{array}{ccc}
-2 & -1 & 7 \\
4 & -2 & 3 \\
3 & 4 & 0
\end{array} \right|$

Now, adding $\displaystyle \frac{-4}{3}$ times column 1 to column 2, we get


$
\begin{equation}
\begin{aligned}

\det (A) =& -7 \left| \begin{array}{ccc}
-2 & \displaystyle \frac{5}{3} & 7 \\
4 & \displaystyle \frac{-22}{3} & 3 \\
3 & 0 & 0
\end{array} \right|
\qquad \text{Expand this by column 1}
\\
\\
=& -7 (3) \left| \begin{array}{cc}
\displaystyle \frac{5}{3} & 7 \\
\displaystyle \frac{-22}{3} & 3
\end{array} \right|
\\
\\
=& -7 (3) \left[ \frac{5}{3} \cdot 3 + \frac{22}{3} \cdot 7 \right]
\\
\\
=& -1183


\end{aligned}
\end{equation}
$