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

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

Evaluate

$$-12$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$4\times 3-2=10$$

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

$$-2-10$$

Subtract 10 from -2.

$$-12$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$2\left(-2\right)-\left(-4\right)-12$$

Add the terms to obtain the final result.

$$-12$$

Factor

$$-12$$