###### 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

• $$(-1)^{i+j}$$ where $$i$$ and $$j$$ are the row and column numbers, respectively, of the entry,

• the entry itself, and

• the determinant of the entries left over when we have crossed out the row and column containing the entry.

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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