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

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

Evaluate

$$20$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2+4\times 2\times 2=14$$

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

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

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

$$14-\left(-6\right)$$

Subtract -6 from 14.

$$20$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$20$$

Factor

$$2^{2}\times 5$$