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

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

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3-2\times 2\times 3+4\left(-2\right)=-17$$

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

$$-17-\left(-17\right)$$

Subtract -17 from -17.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\times 5-\left(-2\left(-2\right)\right)-11$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$