Solve {ccc}{3}&{2}&{-1}{-5}&{5}&{6}{0}&{-1}&{4}

$$\left| \begin{array} { c c c } { 3 } & { 2 } & { - 1 } \\ { - 5 } & { 5 } & { 6 } \\ { 0 } & { - 1 } & { 4 } \end{array} \right|$$

Evaluate

$$113$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&2&-1\\-5&5&6\\0&-1&4\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&2&-1&3&2\\-5&5&6&-5&5\\0&-1&4&0&-1\end{matrix}\right)$$

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

$$3\times 5\times 4-\left(-5\left(-1\right)\right)=55$$

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

$$-6\times 3+4\left(-5\right)\times 2=-58$$

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

$$55-\left(-58\right)$$

Subtract -58 from 55.

$$113$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&2&-1\\-5&5&6\\0&-1&4\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}5&6\\-1&4\end{matrix}\right))-2det(\left(\begin{matrix}-5&6\\0&4\end{matrix}\right))-det(\left(\begin{matrix}-5&5\\0&-1\end{matrix}\right))$$

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

$$3\left(5\times 4-\left(-6\right)\right)-2\left(-5\right)\times 4-\left(-5\left(-1\right)\right)$$

Simplify.

$$3\times 26-2\left(-20\right)-5$$

Add the terms to obtain the final result.

$$113$$

Factor

$$113$$