Solve {lll}{2}&{-5}&{1}{1}&{0}&{-6}{1}&{-7}&{6}

$$\left| \begin{array} { l l l } { 2 } & { - 5 } & { 1 } \\ { 1 } & { 0 } & { - 6 } \\ { 1 } & { - 7 } & { 6 } \end{array} \right|$$

Evaluate

$$-31$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-5&1\\1&0&-6\\1&-7&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-5&1&2&-5\\1&0&-6&1&0\\1&-7&6&1&-7\end{matrix}\right)$$

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

$$-5\left(-6\right)-7=23$$

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

$$-7\left(-6\right)\times 2+6\left(-5\right)=54$$

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

$$23-54$$

Subtract 54 from 23.

$$-31$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-5&1\\1&0&-6\\1&-7&6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}0&-6\\-7&6\end{matrix}\right))-\left(-5det(\left(\begin{matrix}1&-6\\1&6\end{matrix}\right))\right)+det(\left(\begin{matrix}1&0\\1&-7\end{matrix}\right))$$

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

$$2\left(-\left(-7\left(-6\right)\right)\right)-\left(-5\left(6-\left(-6\right)\right)\right)-7$$

Simplify.

$$2\left(-42\right)-\left(-5\times 12\right)-7$$

Add the terms to obtain the final result.

$$-31$$

Factor

$$-31$$