Solve {ccc}{1}&{2}&{1}{2}&{1}&{(-3)}{(-1)}&{1}&{(-1)}

$$\left| \begin{array} { c c c } { 1 } & { 2 } & { 1 } \\ { 2 } & { 1 } & { ( - 3 ) } \\ { ( - 1 ) } & { 1 } & { ( - 1 ) } \end{array} \right|$$

Evaluate

$$15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-1+2\left(-3\right)\left(-1\right)+2=7$$

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

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

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

$$7-\left(-8\right)$$

Subtract -8 from 7.

$$15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2-2\left(-5\right)+3$$

Add the terms to obtain the final result.

$$15$$

Factor

$$3\times 5$$