Solve {rrr}{2}&{1}&{-2}{3}&{-1}&{0}{4}&{1}&{-3}

$$\left| \begin{array} { r r r } { 2 } & { 1 } & { - 2 } \\ { 3 } & { - 1 } & { 0 } \\ { 4 } & { 1 } & { - 3 } \end{array} \right|$$

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&1&-2\\3&-1&0\\4&1&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&1&-2&2&1\\3&-1&0&3&-1\\4&1&-3&4&1\end{matrix}\right)$$

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

$$2\left(-1\right)\left(-3\right)-2\times 3=0$$

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

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

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

$$-\left(-1\right)$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&1&-2\\3&-1&0\\4&1&-3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}-1&0\\1&-3\end{matrix}\right))-det(\left(\begin{matrix}3&0\\4&-3\end{matrix}\right))-2det(\left(\begin{matrix}3&-1\\4&1\end{matrix}\right))$$

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

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

Simplify.

$$2\times 3-\left(-9\right)-2\times 7$$

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$