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

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

Calculate Determinant

$$2$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 2+1+3\times 2=11$$

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

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

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

$$11-9$$

Subtract 9 from 11.

$$2$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2+4$$

Add the terms to obtain the final result.

$$2$$

Evaluate

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