Solve {ccc}{3}&{6}&{9}{2}&{4}&{6}{1}&{2}&{0}

$$\left| \begin{array} { c c c } { 3 } & { 6 } & { 9 } \\ { 2 } & { 4 } & { 6 } \\ { 1 } & { 2 } & { 0 } \end{array} \right|$$

Evaluate

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Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&6&9\\2&4&6\\1&2&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&6&9&3&6\\2&4&6&2&4\\1&2&0&1&2\end{matrix}\right)$$

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

$$6\times 6+9\times 2\times 2=72$$

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

$$4\times 9+2\times 6\times 3=72$$

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

$$72-72$$

Subtract 72 from 72.

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Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&6&9\\2&4&6\\1&2&0\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}4&6\\2&0\end{matrix}\right))-6det(\left(\begin{matrix}2&6\\1&0\end{matrix}\right))+9det(\left(\begin{matrix}2&4\\1&2\end{matrix}\right))$$

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

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

Simplify.

$$3\left(-12\right)-6\left(-6\right)$$

Add the terms to obtain the final result.

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Factor

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