Solve {ccc}{5}&{0}&{0}{0}&{-6}&{-12}{0}&{-12}&{1}

$$\left| \begin{array} { c c c } { 5 } & { 0 } & { 0 } \\ { 0 } & { - 6 } & { - 12 } \\ { 0 } & { - 12 } & { 1 } \end{array} \right|$$

Evaluate

$$-750$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&0&0\\0&-6&-12\\0&-12&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&0&0&5&0\\0&-6&-12&0&-6\\0&-12&1&0&-12\end{matrix}\right)$$

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

$$5\left(-6\right)=-30$$

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

$$-12\left(-12\right)\times 5=720$$

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

$$-30-720$$

Subtract 720 from -30.

$$-750$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&0&0\\0&-6&-12\\0&-12&1\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}-6&-12\\-12&1\end{matrix}\right))$$

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

$$5\left(-6-\left(-12\left(-12\right)\right)\right)$$

Simplify.

$$5\left(-150\right)$$

Add the terms to obtain the final result.

$$-750$$

Factor

$$-750$$