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

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

Evaluate

$$23$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\left(-2\right)\times 6+3=27$$

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

$$6-2=4$$

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

$$27-4$$

Subtract 4 from 27.

$$23$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-\left(-2\left(1-6\left(-2\right)\right)\right)+3-6$$

Simplify.

$$-\left(-2\times 13\right)-3$$

Add the terms to obtain the final result.

$$23$$

Factor

$$23$$