College Algebra, Chapter 7, 7.4, Section 7.4, Problem 30

Find the determinant of the matrix $\displaystyle A = \left| \begin{array}{cccc}
2 & -1 & 6 & 4 \\
7 & 2 & -2 & 5 \\
4 & -2 & 10 & 8 \\
6 & 1 & 1 & 4
\end{array} \right|$, using row/column operations.

If we add 4 times column 2 to column 4, we get

$\displaystyle \left| \begin{array}{cccc}
2 & -1 & 6 & 0 \\
7 & 2 & -2 & 13 \\
4 & -2 & 10 & 0 \\
6 & 1 & 1 & 8
\end{array} \right|$

So,

$\displaystyle \det (A) = 13 \left| \begin{array}{ccc}
2 & -1 & 6 \\
4 & -2 & 10 \\
6 & 1 & 1
\end{array} \right| - 8 \left| \begin{array}{ccc}
2 & -1 & 6 \\
7 & 2 & -2 \\
4 & -2 & 10
\end{array} \right|$

Now, we add $\displaystyle \frac{-1}{2}$ times row 2 to row 1 in the first matrix and add 2 times column 2 to column 1. This gives us


$
\begin{equation}
\begin{aligned}

\det (A) =& 13 \left| \begin{array}{ccc}
0 & 0 & 1 \\
4 & -2 & 10 \\
6 & 1 & 1
\end{array} \right| - 8 \left| \begin{array}{ccc}
0 & -1 & 6 \\
11 & 2 & -2 \\
0 & -2 & 10
\end{array} \right|
\qquad \text{Expand}
\\
\\
=& 13 (1) \left| \begin{array}{cc}
4 & -2 \\
6 & 1
\end{array} \right|
-8 (11) \left| \begin{array}{cc}
-1 & 6 \\
-2 & 10
\end{array} \right|
\\
\\
=& 13(1) \left[ 4 \cdot 1 - (-2) \cdot 6 \right] - 8(11) \left[ (-1) \cdot 10 - 6 \cdot (-2) \right]
\\
\\
=& 208 - 176
\\
\\
=& 32


\end{aligned}
\end{equation}
$

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