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

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

Evaluate

$$2$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

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

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

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

Subtract -3 from -1.

$$2$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$2$$

Factor

$$2$$