Solve {ccc}{2}&{-5}&{1}{1}&{0}&{-6}{0}&{-1}&{2}

$$\left| \begin{array} { c c c } { 2 } & { - 5 } & { 1 } \\ { 1 } & { 0 } & { - 6 } \\ { 0 } & { - 1 } & { 2 } \end{array} \right|$$

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-5&1\\1&0&-6\\0&-1&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-5&1&2&-5\\1&0&-6&1&0\\0&-1&2&0&-1\end{matrix}\right)$$

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

$$-1=-1$$

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

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

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

$$-1-2$$

Subtract 2 from -1.

$$-3$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-5&1\\1&0&-6\\0&-1&2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}0&-6\\-1&2\end{matrix}\right))-\left(-5det(\left(\begin{matrix}1&-6\\0&2\end{matrix}\right))\right)+det(\left(\begin{matrix}1&0\\0&-1\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-3$$

Factor

$$-3$$