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

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

Calculate Determinant

$$48$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\times 4\times 3+3\left(-2\right)\left(-4\right)=48$$

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

$$\text{true}$$

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

$$48$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-2\left(-12\right)+3\times 8$$

Add the terms to obtain the final result.

$$48$$

Evaluate

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