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

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

Evaluate

$$12$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 3-2\times 3=0$$

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

$$2\times 2\left(-2\right)-4=-12$$

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

$$-\left(-12\right)$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$10-2\left(-1\right)$$

Add the terms to obtain the final result.

$$12$$

Factor

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