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

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

Evaluate

$$94$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$6\times 4-4\left(-5\right)\left(-2\right)=-16$$

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

$$4\times 5\left(-4\right)-5\times 6=-110$$

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

$$-16-\left(-110\right)$$

Subtract -110 from -16.

$$94$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-6\left(-9\right)-4\left(-10\right)$$

Add the terms to obtain the final result.

$$94$$

Factor

$$2\times 47$$