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

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

Calculate Determinant

$$24$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2\times 2+2\times 3=10$$

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

$$34-10$$

Subtract 10 from 34.

$$24$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$24$$

Evaluate

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