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

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

Calculate Determinant

$$20$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3+2\times 2+3\times 4\times 3=43$$

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

$$3\times 3+3\times 2+4\times 2=23$$

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

$$43-23$$

Subtract 23 from 43.

$$20$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3-2\times 2+3\times 9$$

Add the terms to obtain the final result.

$$20$$

Evaluate

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