Solve {rrr}{2}&{-1}&{-1}{0}&{-1}&{-3}{0}&{2}&{5}

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

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&-1\\0&-1&-3\\0&2&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&-1&2&-1\\0&-1&-3&0&-1\\0&2&5&0&2\end{matrix}\right)$$

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

$$2\left(-1\right)\times 5=-10$$

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

$$2\left(-3\right)\times 2=-12$$

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

$$-10-\left(-12\right)$$

Subtract -12 from -10.

$$2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&-1\\0&-1&-3\\0&2&5\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-1&-3\\2&5\end{matrix}\right))-\left(-det(\left(\begin{matrix}0&-3\\0&5\end{matrix}\right))\right)-det(\left(\begin{matrix}0&-1\\0&2\end{matrix}\right))$$

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

$$2\left(-5-2\left(-3\right)\right)$$

Simplify.

$$2$$

Factor

$$2$$