Solve {ccc}{2}&{3}&{-5}{0}&{-1}&{-1}{1}&{-2}&{-7}=

$$\left| \begin{array} { c c c } { 2 } & { 3 } & { - 5 } \\ { 0 } & { - 1 } & { - 1 } \\ { 1 } & { - 2 } & { - 7 } \end{array} \right| =$$

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&3&-5\\0&-1&-1\\1&-2&-7\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&3&-5&2&3\\0&-1&-1&0&-1\\1&-2&-7&1&-2\end{matrix}\right)$$

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

$$2\left(-1\right)\left(-7\right)+3\left(-1\right)=11$$

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

$$-\left(-5\right)-2\left(-1\right)\times 2=9$$

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

$$11-9$$

Subtract 9 from 11.

$$2$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&3&-5\\0&-1&-1\\1&-2&-7\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-1&-1\\-2&-7\end{matrix}\right))-3det(\left(\begin{matrix}0&-1\\1&-7\end{matrix}\right))-5det(\left(\begin{matrix}0&-1\\1&-2\end{matrix}\right))$$

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

$$2\left(-\left(-7\right)-\left(-2\left(-1\right)\right)\right)-3\left(-\left(-1\right)\right)-5\left(-\left(-1\right)\right)$$

Simplify.

$$2\times 5-3-5$$

Add the terms to obtain the final result.

$$2$$

Factor

$$2$$