Solve {ccc}{0}&{-5}&{-1}{5}&{0}&{0}{-2}&{1}&{1}

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

Evaluate

$$20$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-5=-5$$

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

$$5\left(-5\right)=-25$$

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

$$-5-\left(-25\right)$$

Subtract -25 from -5.

$$20$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Add the terms to obtain the final result.

$$20$$

Factor

$$2^{2}\times 5$$