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

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

Evaluate

$$36$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2-2\left(-1\right)\left(-4\right)+3\times 5=11$$

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

$$-4\times 2+5\left(-1\right)+2\times 3\left(-2\right)=-25$$

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

$$11-\left(-25\right)$$

Subtract -25 from 11.

$$36$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$9-\left(-2\times 2\right)+23$$

Add the terms to obtain the final result.

$$36$$

Factor

$$2^{2}\times 3^{2}$$