Solve {ccc}{-3}&{0}&{6}{5}&{4}&{3}{2}&{-1}&{1}=

$$\left| \begin{array} { c c c } { - 3 } & { 0 } & { 6 } \\ { 5 } & { 4 } & { 3 } \\ { 2 } & { - 1 } & { 1 } \end{array} \right| =$$

Evaluate

$$-99$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&0&6\\5&4&3\\2&-1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&0&6&-3&0\\5&4&3&5&4\\2&-1&1&2&-1\end{matrix}\right)$$

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

$$-3\times 4+6\times 5\left(-1\right)=-42$$

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

$$2\times 4\times 6-3\left(-3\right)=57$$

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

$$-42-57$$

Subtract 57 from -42.

$$-99$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&0&6\\5&4&3\\2&-1&1\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}4&3\\-1&1\end{matrix}\right))+6det(\left(\begin{matrix}5&4\\2&-1\end{matrix}\right))$$

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

$$-3\left(4-\left(-3\right)\right)+6\left(5\left(-1\right)-2\times 4\right)$$

Simplify.

$$-3\times 7+6\left(-13\right)$$

Add the terms to obtain the final result.

$$-99$$

Factor

$$-99$$