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

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

Calculate Determinant

$$7$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\left(-1\right)\left(-2\right)+2=0$$

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

$$-3+4\left(-1\right)=-7$$

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

$$-\left(-7\right)$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$7$$

Evaluate

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