Solve {rrr}{4}&{0}&{0}{3}&{9}&{-9}{8}&{8}&{-4}

$$\left| \begin{array} { r r r } { 4 } & { 0 } & { 0 } \\ { 3 } & { 9 } & { - 9 } \\ { 8 } & { 8 } & { - 4 } \end{array} \right|$$

Evaluate

$$144$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&0&0\\3&9&-9\\8&8&-4\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&0&0&4&0\\3&9&-9&3&9\\8&8&-4&8&8\end{matrix}\right)$$

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

$$4\times 9\left(-4\right)=-144$$

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

$$8\left(-9\right)\times 4=-288$$

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

$$-144-\left(-288\right)$$

Subtract -288 from -144.

$$144$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&0&0\\3&9&-9\\8&8&-4\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}9&-9\\8&-4\end{matrix}\right))$$

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

$$4\left(9\left(-4\right)-8\left(-9\right)\right)$$

Simplify.

$$4\times 36$$

Add the terms to obtain the final result.

$$144$$

Factor

$$2^{4}\times 3^{2}$$