Solve {lll}{9}&{1}&{0}{6}&{4}&{0}{0}&{3}&{4}

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

Evaluate

$$120$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$9\times 4\times 4=144$$

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

$$4\times 6=24$$

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

$$144-24$$

Subtract 24 from 144.

$$120$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$9\times 4\times 4-6\times 4$$

Simplify.

$$9\times 16-24$$

Add the terms to obtain the final result.

$$120$$

Factor

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