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

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

Evaluate

$$42$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3\times 2+2\times 4\times 5=34$$

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

$$3\times 4\left(-1\right)+2\times 2=-8$$

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

$$34-\left(-8\right)$$

Subtract -8 from 34.

$$42$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-6\right)-2\left(-18\right)$$

Add the terms to obtain the final result.

$$42$$

Factor

$$2\times 3\times 7$$