Solve {ccc}{2}&{5}&{3}{0}&{4}&{7}{-2}&{-2}&{3}

$$\left| \begin{array} { c c c } { 2 } & { 5 } & { 3 } \\ { 0 } & { 4 } & { 7 } \\ { - 2 } & { - 2 } & { 3 } \end{array} \right|$$

Evaluate

$$6$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&5&3\\0&4&7\\-2&-2&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&5&3&2&5\\0&4&7&0&4\\-2&-2&3&-2&-2\end{matrix}\right)$$

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

$$2\times 4\times 3+5\times 7\left(-2\right)=-46$$

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

$$-2\times 4\times 3-2\times 7\times 2=-52$$

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

$$-46-\left(-52\right)$$

Subtract -52 from -46.

$$6$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&5&3\\0&4&7\\-2&-2&3\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}4&7\\-2&3\end{matrix}\right))-5det(\left(\begin{matrix}0&7\\-2&3\end{matrix}\right))+3det(\left(\begin{matrix}0&4\\-2&-2\end{matrix}\right))$$

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

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

Simplify.

$$2\times 26-5\times 14+3\times 8$$

Add the terms to obtain the final result.

$$6$$

Factor

$$2\times 3$$