Question

Evaluate each determinant using expansion by minors.

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

Short answer

The value of the determinant $\left|\begin{array}{ccc}0& -4& 0\\ 3& -2& 5\\ 2& -1& 1\end{array}\right|=-28$.

Step-by-step solution

- Decide row to use for expansion

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

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

- 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.

localid="1648141193476" $\begin{array}{c}\left|\begin{array}{ccc}0& -4& 0\\ 3& -2& 5\\ 2& -1& 1\end{array}\right|=0\left|\begin{array}{cc}-2& 5\\ -1& 1\end{array}\right|-\left(-4\right)\left|\begin{array}{cc}3& 5\\ 2& 1\end{array}\right|+0\left|\begin{array}{cc}3& -2\\ 2& -1\end{array}\right|\\ =0\left(-2-\left(-5\right)\right)-\left(-4\right)\left(3-10\right)+0\left(-3-\left(-2\right)\right)\end{array}$

- Multiply and simplify

Further simplify the values.

$\begin{array}{c}\left|\begin{array}{ccc}0& -4& 0\\ 3& -2& 5\\ 2& -1& 1\end{array}\right|=0\left(3\right)-\left(-4\right)\left(-7\right)+0\left(-1\right)\\ =0+4\left(-7\right)+0\\ =-28\end{array}$