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

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

Calculate Determinant

$$5$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 4+2\times 3\left(-1\right)=-14$$

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

$$-1+3\times 3\left(-2\right)=-19$$

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

$$-14-\left(-19\right)$$

Subtract -19 from -14.

$$5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$5$$

Evaluate

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