Solve {ccc}{3}&{-1}&{2}{-1}&{6}&{0}{4}&{5}&{-2}

$$\left| \begin{array} { c c c } { 3 } & { - 1 } & { 2 } \\ { - 1 } & { 6 } & { 0 } \\ { 4 } & { 5 } & { - 2 } \end{array} \right|$$

Evaluate

$$-92$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&-1&2\\-1&6&0\\4&5&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&-1&2&3&-1\\-1&6&0&-1&6\\4&5&-2&4&5\end{matrix}\right)$$

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

$$3\times 6\left(-2\right)+2\left(-1\right)\times 5=-46$$

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

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

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

$$-46-46$$

Subtract 46 from -46.

$$-92$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&-1&2\\-1&6&0\\4&5&-2\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}6&0\\5&-2\end{matrix}\right))-\left(-det(\left(\begin{matrix}-1&0\\4&-2\end{matrix}\right))\right)+2det(\left(\begin{matrix}-1&6\\4&5\end{matrix}\right))$$

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

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

Simplify.

$$3\left(-12\right)-\left(-2\right)+2\left(-29\right)$$

Add the terms to obtain the final result.

$$-92$$

Factor

$$-92$$