Solve {lll}{1}&{2}&{1}{1}&{2}&{2}{1}&{1}&{5}

$$\left| \begin{array} { l l l } { 1 } & { 2 } & { 1 } \\ { 1 } & { 2 } & { 2 } \\ { 1 } & { 1 } & { 5 } \end{array} \right|$$

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&2&1\\1&2&2\\1&1&5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&2&1&1&2\\1&2&2&1&2\\1&1&5&1&1\end{matrix}\right)$$

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

$$2\times 5+2\times 2+1=15$$

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

$$2+2+5\times 2=14$$

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

$$15-14$$

Subtract 14 from 15.

$$1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&2&1\\1&2&2\\1&1&5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}2&2\\1&5\end{matrix}\right))-2det(\left(\begin{matrix}1&2\\1&5\end{matrix}\right))+det(\left(\begin{matrix}1&2\\1&1\end{matrix}\right))$$

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

$$2\times 5-2-2\left(5-2\right)+1-2$$

Simplify.

$$8-2\times 3-1$$

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$