Solve {lll}{0}&{2}&{0}{1}&{-1}&{0}{0}&{0}&{-4}

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

Evaluate

$$8$$

Steps Using Sarrus' Rule

Find the determinant of the matrix using the method of diagonals.

$$det(\left(\begin{matrix}0&2&0\\1&-1&0\\0&0&-4\end{matrix}\right))$$

Extend the original matrix by repeating the first two columns as the fourth and fifth columns.

$$\left(\begin{matrix}0&2&0&0&2\\1&-1&0&1&-1\\0&0&-4&0&0\end{matrix}\right)$$

Starting at the upper left entry, multiply down along the diagonals, and add the resulting products.

$$\text{true}$$

Starting at the lower left entry, multiply up along the diagonals, and add the resulting products.

$$-4\times 2=-8$$

Subtract the sum of the upward diagonal products from the sum of the downward diagonal products.

$$-\left(-8\right)$$

Steps Using Expansion by Minors

Find the determinant of the matrix using the method of expansion by minors (also known as expansion by cofactors).

$$det(\left(\begin{matrix}0&2&0\\1&-1&0\\0&0&-4\end{matrix}\right))$$

To expand by minors, multiply each element of the first row by its minor, which is the determinant of the 2\times 2 matrix created by deleting the row and column containing that element, then multiply by the element's position sign.

$$-2det(\left(\begin{matrix}1&0\\0&-4\end{matrix}\right))$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the determinant is ad-bc.

$$-2\left(-4\right)$$

Add the terms to obtain the final result.

$$8$$

Factor

$$2^{3}$$