Solve {ccc}{5}&{2}&{1}{3}&{1}&{3}{-4}&{-1}&{-3}=

$$\left| \begin{array} { c c c } { 5 } & { 2 } & { 1 } \\ { 3 } & { 1 } & { 3 } \\ { - 4 } & { - 1 } & { - 3 } \end{array} \right| =$$

Evaluate

$$-5$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&2&1\\3&1&3\\-4&-1&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&2&1&5&2\\3&1&3&3&1\\-4&-1&-3&-4&-1\end{matrix}\right)$$

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

$$5\left(-3\right)+2\times 3\left(-4\right)+3\left(-1\right)=-42$$

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

$$-4-3\times 5-3\times 3\times 2=-37$$

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

$$-42-\left(-37\right)$$

Subtract -37 from -42.

$$-5$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&2&1\\3&1&3\\-4&-1&-3\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}1&3\\-1&-3\end{matrix}\right))-2det(\left(\begin{matrix}3&3\\-4&-3\end{matrix}\right))+det(\left(\begin{matrix}3&1\\-4&-1\end{matrix}\right))$$

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

$$5\left(-3-\left(-3\right)\right)-2\left(3\left(-3\right)-\left(-4\times 3\right)\right)+3\left(-1\right)-\left(-4\right)$$

Simplify.

$$-2\times 3+1$$

Add the terms to obtain the final result.

$$-5$$

Factor

$$-5$$