Solve {rrr}{2}&{-1}&{3}{2}&{2}&{4}{1}&{-8}&{-6}

$$\left| \begin{array} { r r r } { 2 } & { - 1 } & { 3 } \\ { 2 } & { 2 } & { 4 } \\ { 1 } & { - 8 } & { - 6 } \end{array} \right|$$

Evaluate

$$-30$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&3\\2&2&4\\1&-8&-6\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&3&2&-1\\2&2&4&2&2\\1&-8&-6&1&-8\end{matrix}\right)$$

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

$$2\times 2\left(-6\right)-4+3\times 2\left(-8\right)=-76$$

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

$$2\times 3-8\times 4\times 2-6\times 2\left(-1\right)=-46$$

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

$$-76-\left(-46\right)$$

Subtract -46 from -76.

$$-30$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&3\\2&2&4\\1&-8&-6\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}2&4\\-8&-6\end{matrix}\right))-\left(-det(\left(\begin{matrix}2&4\\1&-6\end{matrix}\right))\right)+3det(\left(\begin{matrix}2&2\\1&-8\end{matrix}\right))$$

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

$$2\left(2\left(-6\right)-\left(-8\times 4\right)\right)-\left(-\left(2\left(-6\right)-4\right)\right)+3\left(2\left(-8\right)-2\right)$$

Simplify.

$$2\times 20-\left(-\left(-16\right)\right)+3\left(-18\right)$$

Add the terms to obtain the final result.

$$-30$$

Factor

$$-30$$