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

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

Calculate Determinant

$$4$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 2-\left(-2\right)+3\times 4=10$$

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

$$-2\left(-2\right)+4\left(-1\right)+2\times 3=6$$

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

$$10-6$$

Subtract 6 from 10.

$$4$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4+8$$

Add the terms to obtain the final result.

$$4$$

Evaluate

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