Solve {ccc}{1}&{3}&{4}{9}&{15}&{0}{-6}&{0}&{24}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 4 } \\ { 9 } & { 15 } & { 0 } \\ { - 6 } & { 0 } & { 24 } \end{array} \right|$$

Evaluate

$$72$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&4\\9&15&0\\-6&0&24\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&4&1&3\\9&15&0&9&15\\-6&0&24&-6&0\end{matrix}\right)$$

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

$$15\times 24=360$$

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

$$-6\times 15\times 4+24\times 9\times 3=288$$

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

$$360-288$$

Subtract 288 from 360.

$$72$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&4\\9&15&0\\-6&0&24\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}15&0\\0&24\end{matrix}\right))-3det(\left(\begin{matrix}9&0\\-6&24\end{matrix}\right))+4det(\left(\begin{matrix}9&15\\-6&0\end{matrix}\right))$$

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

$$15\times 24-3\times 9\times 24+4\left(-\left(-6\times 15\right)\right)$$

Simplify.

$$360-3\times 216+4\times 90$$

Add the terms to obtain the final result.

$$72$$

Factor

$$2^{3}\times 3^{2}$$