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

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

Evaluate

$$18$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$4\times 2\left(-2\right)=-16$$

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

$$2-\left(-16\right)$$

Subtract -16 from 2.

$$18$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-4-\left(-2\times 11\right)$$

Add the terms to obtain the final result.

$$18$$

Factor

$$2\times 3^{2}$$