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

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

Evaluate

$$-29$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\left(-1\right)\left(-3\right)+2\left(-3\right)\left(-1\right)=15$$

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

$$-14-15$$

Subtract 15 from -14.

$$-29$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2-2\times 9-3\times 3$$

Add the terms to obtain the final result.

$$-29$$

Factor

$$-29$$