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

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

Evaluate

$$-104$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4\left(-3\right)+5\times 6\left(-1\right)=-18$$

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

$$4\times 4\times 5-6\left(-1\right)=86$$

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

$$-18-86$$

Subtract 86 from -18.

$$-104$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-6\right)+5\left(-22\right)$$

Add the terms to obtain the final result.

$$-104$$

Factor

$$-104$$