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

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

Evaluate

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Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$23-23$$

Subtract 23 from 23.

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Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$7-2\times 5+3$$

Add the terms to obtain the final result.

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Factor

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