Solve {rrr}{2}&{-1}&{1}{-1}&{4}&{1}{1}&{1}&{3}

$$\left| \begin{array} { r r r } { 2 } & { - 1 } & { 1 } \\ { - 1 } & { 4 } & { 1 } \\ { 1 } & { 1 } & { 3 } \end{array} \right|$$

Evaluate

$$13$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&1\\-1&4&1\\1&1&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&1&2&-1\\-1&4&1&-1&4\\1&1&3&1&1\end{matrix}\right)$$

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

$$2\times 4\times 3-1-1=22$$

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

$$4+2+3\left(-1\right)\left(-1\right)=9$$

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

$$22-9$$

Subtract 9 from 22.

$$13$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&1\\-1&4&1\\1&1&3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}4&1\\1&3\end{matrix}\right))-\left(-det(\left(\begin{matrix}-1&1\\1&3\end{matrix}\right))\right)+det(\left(\begin{matrix}-1&4\\1&1\end{matrix}\right))$$

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

$$2\left(4\times 3-1\right)-\left(-\left(-3-1\right)\right)-1-4$$

Simplify.

$$2\times 11-\left(-\left(-4\right)\right)-5$$

Add the terms to obtain the final result.

$$13$$

Factor

$$13$$