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

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

Evaluate

$$-27$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\times 5-1=-16$$

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

$$5-2\left(-3\right)=11$$

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

$$-16-11$$

Subtract 11 from -16.

$$-27$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-27$$

Factor

$$-27$$