Solve {rrr}{1}&{-1}&{2}{2}&{-5}&{-5}{1}&{-7}&{3}

$$\left| \begin{array} { r r r } { 1 } & { - 1 } & { 2 } \\ { 2 } & { - 5 } & { - 5 } \\ { 1 } & { - 7 } & { 3 } \end{array} \right|$$

Evaluate

$$-57$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&-1&2\\2&-5&-5\\1&-7&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&-1&2&1&-1\\2&-5&-5&2&-5\\1&-7&3&1&-7\end{matrix}\right)$$

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

$$-5\times 3-\left(-5\right)+2\times 2\left(-7\right)=-38$$

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

$$-5\times 2-7\left(-5\right)+3\times 2\left(-1\right)=19$$

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

$$-38-19$$

Subtract 19 from -38.

$$-57$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&-1&2\\2&-5&-5\\1&-7&3\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-5&-5\\-7&3\end{matrix}\right))-\left(-det(\left(\begin{matrix}2&-5\\1&3\end{matrix}\right))\right)+2det(\left(\begin{matrix}2&-5\\1&-7\end{matrix}\right))$$

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

$$-5\times 3-\left(-7\left(-5\right)\right)-\left(-\left(2\times 3-\left(-5\right)\right)\right)+2\left(2\left(-7\right)-\left(-5\right)\right)$$

Simplify.

$$-50-\left(-11\right)+2\left(-9\right)$$

Add the terms to obtain the final result.

$$-57$$

Factor

$$-57$$