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

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

Evaluate

$$25$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-3+2\times 4\times 3=21$$

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

$$-2\times 2=-4$$

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

$$21-\left(-4\right)$$

Subtract -4 from 21.

$$25$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-3-2\left(-14\right)$$

Add the terms to obtain the final result.

$$25$$

Factor

$$5^{2}$$