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

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

Evaluate

$$0$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 4+2\times 5\times 5+3\left(-2\right)\times 3=44$$

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

$$5\times 3\times 3+3\times 5+4\left(-2\right)\times 2=44$$

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

$$44-44$$

Subtract 44 from 44.

$$0$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3-2\left(-33\right)+3\left(-21\right)$$

Add the terms to obtain the final result.

$$0$$

Factor

$$0$$