Solve {rrr}{3}&{0}&{-16}{4}&{5}&{-16}{0}&{6}&{-9}

$$\left| \begin{array} { r r r } { 3 } & { 0 } & { - 16 } \\ { 4 } & { 5 } & { - 16 } \\ { 0 } & { 6 } & { - 9 } \end{array} \right|$$

Evaluate

$$-231$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&0&-16\\4&5&-16\\0&6&-9\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&0&-16&3&0\\4&5&-16&4&5\\0&6&-9&0&6\end{matrix}\right)$$

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

$$3\times 5\left(-9\right)-16\times 4\times 6=-519$$

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

$$6\left(-16\right)\times 3=-288$$

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

$$-519-\left(-288\right)$$

Subtract -288 from -519.

$$-231$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&0&-16\\4&5&-16\\0&6&-9\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}5&-16\\6&-9\end{matrix}\right))-16det(\left(\begin{matrix}4&5\\0&6\end{matrix}\right))$$

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

$$3\left(5\left(-9\right)-6\left(-16\right)\right)-16\times 4\times 6$$

Simplify.

$$3\times 51-16\times 24$$

Add the terms to obtain the final result.

$$-231$$

Factor

$$-231$$