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

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

Evaluate

$$-15$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$2\left(-1\right)+4\times 2=6$$

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

$$-9-6$$

Subtract 6 from -9.

$$-15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-2\right)-5-3\times 4$$

Add the terms to obtain the final result.

$$-15$$

Factor

$$-15$$