Solve {ccc}{3}&{0}&{12}{-6}&{3}&{0}{9}&{6}&{3}

$$\left( \begin{array} { c c c } { 3 } & { 0 } & { 12 } \\ { - 6 } & { 3 } & { 0 } \\ { 9 } & { 6 } & { 3 } \end{array} \right)$$

Calculate Determinant

$$-729$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&0&12\\-6&3&0\\9&6&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&0&12&3&0\\-6&3&0&-6&3\\9&6&3&9&6\end{matrix}\right)$$

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

$$3\times 3\times 3+12\left(-6\right)\times 6=-405$$

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

$$9\times 3\times 12=324$$

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

$$-405-324$$

Subtract 324 from -405.

$$-729$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&0&12\\-6&3&0\\9&6&3\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}3&0\\6&3\end{matrix}\right))+12det(\left(\begin{matrix}-6&3\\9&6\end{matrix}\right))$$

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

$$3\times 3\times 3+12\left(-6\times 6-9\times 3\right)$$

Simplify.

$$3\times 9+12\left(-63\right)$$

Add the terms to obtain the final result.

$$-729$$

Evaluate

$$\left(\begin{matrix}3&0&12\\-6&3&0\\9&6&3\end{matrix}\right)$$