Solve {rrr}{3}&{-1}&{1}{4}&{2}&{-1}{6}&{-8}&{5}

$$\left| \begin{array} { r r r } { 3 } & { - 1 } & { 1 } \\ { 4 } & { 2 } & { - 1 } \\ { 6 } & { - 8 } & { 5 } \end{array} \right|$$

Evaluate

$$-12$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&-1&1\\4&2&-1\\6&-8&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&-1&1&3&-1\\4&2&-1&4&2\\6&-8&5&6&-8\end{matrix}\right)$$

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

$$3\times 2\times 5-\left(-6\right)+4\left(-8\right)=4$$

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

$$6\times 2-8\left(-1\right)\times 3+5\times 4\left(-1\right)=16$$

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

$$4-16$$

Subtract 16 from 4.

$$-12$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&-1&1\\4&2&-1\\6&-8&5\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}2&-1\\-8&5\end{matrix}\right))-\left(-det(\left(\begin{matrix}4&-1\\6&5\end{matrix}\right))\right)+det(\left(\begin{matrix}4&2\\6&-8\end{matrix}\right))$$

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

$$3\left(2\times 5-\left(-8\left(-1\right)\right)\right)-\left(-\left(4\times 5-6\left(-1\right)\right)\right)+4\left(-8\right)-6\times 2$$

Simplify.

$$3\times 2-\left(-26\right)-44$$

Add the terms to obtain the final result.

$$-12$$

Factor

$$-12$$