Solve {lll}{2}&{2}&{3}{1}&{-1}&{0}{-1}&{2}&{1}

$$\left( \begin{array} { l l l } { 2 } & { 2 } & { 3 } \\ { 1 } & { - 1 } & { 0 } \\ { - 1 } & { 2 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&2&3\\1&-1&0\\-1&2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&2&3&2&2\\1&-1&0&1&-1\\-1&2&1&-1&2\end{matrix}\right)$$

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

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

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

$$-\left(-1\right)\times 3+2=5$$

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

$$4-5$$

Subtract 5 from 4.

$$-1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&2&3\\1&-1&0\\-1&2&1\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-1&0\\2&1\end{matrix}\right))-2det(\left(\begin{matrix}1&0\\-1&1\end{matrix}\right))+3det(\left(\begin{matrix}1&-1\\-1&2\end{matrix}\right))$$

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-1$$

Evaluate

$$\left(\begin{matrix}2&2&3\\1&-1&0\\-1&2&1\end{matrix}\right)$$