Solve {lll}{1}&{3}&{2}{5}&{2}&{3}{4}&{1}&{0}

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

Evaluate

$$27$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 4+2\times 5=46$$

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

$$4\times 2\times 2+3=19$$

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

$$46-19$$

Subtract 19 from 46.

$$27$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3-3\left(-12\right)+2\left(-3\right)$$

Add the terms to obtain the final result.

$$27$$

Factor

$$3^{3}$$