Solve {ccc}{1}&{2}&{-1}{-1}&{15}&{16}{3}&{1}&{-2}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { - 1 } \\ { - 1 } & { 15 } & { 16 } \\ { 3 } & { 1 } & { - 2 } \end{array} \right|$$

Evaluate

$$92$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&-1\\-1&15&16\\3&1&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&-1&1&2\\-1&15&16&-1&15\\3&1&-2&3&1\end{matrix}\right)$$

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

$$15\left(-2\right)+2\times 16\times 3-\left(-1\right)=67$$

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

$$3\times 15\left(-1\right)+16-2\left(-1\right)\times 2=-25$$

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

$$67-\left(-25\right)$$

Subtract -25 from 67.

$$92$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&-1\\-1&15&16\\3&1&-2\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}15&16\\1&-2\end{matrix}\right))-2det(\left(\begin{matrix}-1&16\\3&-2\end{matrix}\right))-det(\left(\begin{matrix}-1&15\\3&1\end{matrix}\right))$$

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

$$15\left(-2\right)-16-2\left(-\left(-2\right)-3\times 16\right)-\left(-1-3\times 15\right)$$

Simplify.

$$-46-2\left(-46\right)-\left(-46\right)$$

Add the terms to obtain the final result.

$$92$$

Factor

$$2^{2}\times 23$$