Solve {rrr}{4}&{0}&{2}{6}&{-9}&{8}{6}&{-3}&{3}

$$\left| \begin{array} { r r r } { 4 } & { 0 } & { 2 } \\ { 6 } & { - 9 } & { 8 } \\ { 6 } & { - 3 } & { 3 } \end{array} \right|$$

Evaluate

$$60$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&0&2\\6&-9&8\\6&-3&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&0&2&4&0\\6&-9&8&6&-9\\6&-3&3&6&-3\end{matrix}\right)$$

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

$$4\left(-9\right)\times 3+2\times 6\left(-3\right)=-144$$

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

$$6\left(-9\right)\times 2-3\times 8\times 4=-204$$

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

$$-144-\left(-204\right)$$

Subtract -204 from -144.

$$60$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&0&2\\6&-9&8\\6&-3&3\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}-9&8\\-3&3\end{matrix}\right))+2det(\left(\begin{matrix}6&-9\\6&-3\end{matrix}\right))$$

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

$$4\left(-9\times 3-\left(-3\times 8\right)\right)+2\left(6\left(-3\right)-6\left(-9\right)\right)$$

Simplify.

$$4\left(-3\right)+2\times 36$$

Add the terms to obtain the final result.

$$60$$

Factor

$$2^{2}\times 3\times 5$$