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

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

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-1-3\times 3=-10$$

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

$$3\times 2-\left(-\left(-1\right)\right)=5$$

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

$$-10-5$$

Subtract 5 from -10.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-1-\left(-\left(-8\right)\right)+2\left(-3\right)$$

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$