Solve {ccc}{2}&{-2}&{-5}{1}&{6}&{-3}{2}&{1}&{-1}

$$\left( \begin{array} { c c c } { 2 } & { - 2 } & { - 5 } \\ { 1 } & { 6 } & { - 3 } \\ { 2 } & { 1 } & { - 1 } \end{array} \right)$$

Calculate Determinant

$$59$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-2&-5\\1&6&-3\\2&1&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-2&-5&2&-2\\1&6&-3&1&6\\2&1&-1&2&1\end{matrix}\right)$$

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

$$2\times 6\left(-1\right)-2\left(-3\right)\times 2-5=-5$$

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

$$2\times 6\left(-5\right)-3\times 2-\left(-2\right)=-64$$

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

$$-5-\left(-64\right)$$

Subtract -64 from -5.

$$59$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-2&-5\\1&6&-3\\2&1&-1\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}6&-3\\1&-1\end{matrix}\right))-\left(-2det(\left(\begin{matrix}1&-3\\2&-1\end{matrix}\right))\right)-5det(\left(\begin{matrix}1&6\\2&1\end{matrix}\right))$$

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

$$2\left(6\left(-1\right)-\left(-3\right)\right)-\left(-2\left(-1-2\left(-3\right)\right)\right)-5\left(1-2\times 6\right)$$

Simplify.

$$2\left(-3\right)-\left(-2\times 5\right)-5\left(-11\right)$$

Add the terms to obtain the final result.

$$59$$

Evaluate

$$\left(\begin{matrix}2&-2&-5\\1&6&-3\\2&1&-1\end{matrix}\right)$$