Solve {rrr}{-2}&{1}&{1}{1}&{2}&{0}{-1}&{6}&{3}

$$\left| \begin{array} { r r r } { - 2 } & { 1 } & { 1 } \\ { 1 } & { 2 } & { 0 } \\ { - 1 } & { 6 } & { 3 } \end{array} \right|$$

Evaluate

$$-7$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&1&1\\1&2&0\\-1&6&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&1&1&-2&1\\1&2&0&1&2\\-1&6&3&-1&6\end{matrix}\right)$$

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

$$-2\times 2\times 3+6=-6$$

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

$$-2+3=1$$

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

$$-6-1$$

Subtract 1 from -6.

$$-7$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&1&1\\1&2&0\\-1&6&3\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}2&0\\6&3\end{matrix}\right))-det(\left(\begin{matrix}1&0\\-1&3\end{matrix}\right))+det(\left(\begin{matrix}1&2\\-1&6\end{matrix}\right))$$

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

$$-2\times 2\times 3-3+6-\left(-2\right)$$

Simplify.

$$-2\times 6-3+8$$

Add the terms to obtain the final result.

$$-7$$

Factor

$$-7$$