Solve {ccc}{1}&{3}&{-5}{5}&{-2}&{7}{3}&{-4}&{6}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { - 5 } \\ { 5 } & { - 2 } & { 7 } \\ { 3 } & { - 4 } & { 6 } \end{array} \right|$$

Evaluate

$$59$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&-5\\5&-2&7\\3&-4&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&-5&1&3\\5&-2&7&5&-2\\3&-4&6&3&-4\end{matrix}\right)$$

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

$$-2\times 6+3\times 7\times 3-5\times 5\left(-4\right)=151$$

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

$$3\left(-2\right)\left(-5\right)-4\times 7+6\times 5\times 3=92$$

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

$$151-92$$

Subtract 92 from 151.

$$59$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&-5\\5&-2&7\\3&-4&6\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-2&7\\-4&6\end{matrix}\right))-3det(\left(\begin{matrix}5&7\\3&6\end{matrix}\right))-5det(\left(\begin{matrix}5&-2\\3&-4\end{matrix}\right))$$

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

$$-2\times 6-\left(-4\times 7\right)-3\left(5\times 6-3\times 7\right)-5\left(5\left(-4\right)-3\left(-2\right)\right)$$

Simplify.

$$16-3\times 9-5\left(-14\right)$$

Add the terms to obtain the final result.

$$59$$

Factor

$$59$$