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

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

Evaluate

$$-6$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3+2\times 2+2=9$$

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

$$2+2\times 2\times 3+1=15$$

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

$$9-15$$

Subtract 15 from 9.

$$-6$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-3\right)-\left(-3\right)$$

Add the terms to obtain the final result.

$$-6$$

Factor

$$-6$$