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

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

Calculate Determinant

$$-9$$

Steps Using Sarrus' Rule

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

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

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

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

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.

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

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

$$-6-3$$

Subtract 3 from -6.

$$-9$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-5-4$$

Add the terms to obtain the final result.

$$-9$$

Evaluate

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