Solve {rrr}{1}&{0}&{1}{-1}&{-5}&{0}{0}&{-4}&{2}

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&0&1\\-1&-5&0\\0&-4&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&0&1&1&0\\-1&-5&0&-1&-5\\0&-4&2&0&-4\end{matrix}\right)$$

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

$$-5\times 2-\left(-4\right)=-6$$

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

$$\text{true}$$

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

$$-6$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&0&1\\-1&-5&0\\0&-4&2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-5&0\\-4&2\end{matrix}\right))+det(\left(\begin{matrix}-1&-5\\0&-4\end{matrix}\right))$$

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

$$-5\times 2-\left(-4\right)$$

Simplify.

$$-10+4$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$