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

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

Calculate Determinant

$$-5$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

$$-9-\left(-4\right)$$

Subtract -4 from -9.

$$-5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-1-2\left(-8\right)-2\times 10$$

Add the terms to obtain the final result.

$$-5$$

Evaluate

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