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

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

Evaluate

$$-7$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-5\times 3\times 3+4+3\times 2=-35$$

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

$$3\times 3+2\times 4\left(-5\right)+3=-28$$

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

$$-35-\left(-28\right)$$

Subtract -28 from -35.

$$-7$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-7$$

Factor

$$-7$$