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

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

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-2\right)\times 5+2\times 2=-16$$

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

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

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

$$-16-\left(-1\right)$$

Subtract -1 from -16.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-12\right)-\left(-9\right)$$

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$