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

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

Calculate Determinant

$$-10$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 3=6$$

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

$$-2\left(-2\right)+3\left(-2\right)\left(-2\right)=16$$

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

$$6-16$$

Subtract 16 from 6.

$$-10$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2-\left(-2\left(-6\right)\right)$$

Add the terms to obtain the final result.

$$-10$$

Evaluate

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