Example3.4.9

We illustrate how to use a cofactor expansion to find the determinant of \(A\) where

\begin{equation*} A = \left[\begin{array}{rrr} 1 \amp -1 \amp 2 \\ -2 \amp 2 \amp -6 \\ 3 \amp -1 \amp 10 \\ \end{array}\right] \text{.} \end{equation*}

This is the same matrix that appeared in the last activity where we found that \(\det A = 4\text{.}\)

To begin, we choose one row or column. It doesn't matter which we choose because the result will be the same in any case. Here, we will choose the second row.

\begin{equation*} \left[\begin{array}{rrr} \lgray{1} \amp \lgray{-1} \amp \lgray{2} \\ -2 \amp 2 \amp -6 \\ \lgray{3} \amp \lgray{-1} \amp \lgray{10} \\ \end{array}\right] \text{.} \end{equation*}

The determinant will be found by creating a sum of terms, one for each entry in the row we have chosen. For each entry in the row, we will form its term in the cofactor expansion by multiplying

Since we are computing the determinant of this matrix

\begin{equation*} \left[\begin{array}{rrr} \gray{1} \amp \gray{-1} \amp \gray{2} \\ -2 \amp 2 \amp -6 \\ \gray{3} \amp \gray{-1} \amp \gray{10} \\ \end{array}\right] \end{equation*}

using the second row, the entry in the first column of this row is \(-2\text{.}\) Let's see how to form the term from this entry.

The term itself is \(-2\text{,}\) and the matrix that is left over when we cross out the second row and first column is

\begin{equation*} \left[\begin{array}{rrr} \gray{1} \amp {-1} \amp {2} \\ \gray{-2} \amp \gray{2} \amp \gray{-6} \\ \gray{3} \amp {-1} \amp {10} \\ \end{array}\right] \end{equation*}

whose determinant is

\begin{equation*} \det\left[\begin{array}{rr} -1 \amp 2 \\ -1 \amp 10 \\ \end{array}\right] = -1(10) - 2 (-1) = -8 \text{.} \end{equation*}

Since this entry is in the second row and first column, the term we construct is \((-1)^{2+1}(-2)(-8) = -16 \text{.}\)

Putting this together, we find the determinant to be

\begin{equation*} \begin{aligned} \left[\begin{array}{rrr} {1} \amp {-1} \amp {2} \\ -2 \amp {2} \amp {-6} \\ {3} \amp {-1} \amp {10} \\ \end{array}\right] {}={} \amp (-1)^{2+1}(-2)\det\left[\begin{array}{rr} -1 \amp 2 \\ -1 \amp 10 \\ \end{array}\right] \\ \amp {}+{} (-1)^{2+2}(2)\det\left[\begin{array}{rr} 1 \amp 2 \\ 3 \amp 10 \\ \end{array}\right] \\ \amp {}+{} (-1)^{2+3}(-6)\det\left[\begin{array}{rr} -1 \amp -1 \\ 3 \amp -1 \\ \end{array}\right] \\ \\ {}={} \amp (-1)(-2)(-1(10)-2(-1)) \\ \amp + (1)(2)(1(10)-2(3)) \\ \amp + (-1)(-6)((-1)(-1)-(-1)3) \\ \\ {}={} \amp -16 + 8 + 12 \\ {}={} \amp 4 \\ \end{aligned} \text{.} \end{equation*}

Notice that this agrees with the determinant that we found for this matrix using row operations in the last activity.

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