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

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

Evaluate

$$3$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 5-4\left(-2\right)\left(-5\right)+1=-29$$

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

$$-5\times 2-2+5\left(-4\right)=-32$$

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

$$-29-\left(-32\right)$$

Subtract -32 from -29.

$$3$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$12-\left(-4\left(-5\right)\right)+11$$

Add the terms to obtain the final result.

$$3$$

Factor

$$3$$