Question

Evaluate each determinant using expansion by minors.

$\left|\begin{array}{ccc}-3& 0& 6\\ 6& 5& -2\\ 1& 4& 2\end{array}\right|$

Short answer

The value of the determinant $\left|\begin{array}{ccc}-3& 0& 6\\ 6& 5& -2\\ 1& 4& 2\end{array}\right|=60$.

Step-by-step solution

- Decide row to use for expansion

In this case, use first row for the expansion by minors.

$\left|\begin{array}{ccc}-3& 0& 6\\ 6& 5& -2\\ 1& 4& 2\end{array}\right|=-3\left|\begin{array}{cc}5& -2\\ 4& 2\end{array}\right|-\left(0\right)\left|\begin{array}{cc}6& -2\\ 1& 2\end{array}\right|+6\left|\begin{array}{cc}6& 5\\ 1& 4\end{array}\right|$

- Evaluate 2×2 determinants

The determinant of second order matrix is found by calculating the difference of the product of the two diagonals, that is., $\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc$.Apply this definition to find the $2\times 2$determinant.

$\begin{array}{c}\left|\begin{array}{ccc}-3& 0& 6\\ 6& 5& -2\\ 1& 4& 2\end{array}\right|=-3\left|\begin{array}{cc}5& -2\\ 4& 2\end{array}\right|-\left(0\right)\left|\begin{array}{cc}6& -2\\ 1& 2\end{array}\right|+6\left|\begin{array}{cc}6& 5\\ 1& 4\end{array}\right|\\ =-3\left(10-\left(-8\right)\right)-\left(0\right)\left(12-\left(-2\right)\right)+6\left(24-5\right)\end{array}$

- Multiply and simplify

Further simplify the values.

$\begin{array}{c}\left|\begin{array}{ccc}-3& 0& 6\\ 6& 5& -2\\ 1& 4& 2\end{array}\right|=-3\left(10-\left(-8\right)\right)-\left(0\right)\left(12-\left(-2\right)\right)+6\left(24-5\right)\\ =-3\left(18\right)-0\left(14\right)+6\left(19\right)\\ =-54-0+114\\ =60\end{array}$