Solve {ccc}{2}&{4}&{23}{-3}&{6}&{2}{1}&{5}&{-2}

$$\left| \begin{array} { c c c } { 2 } & { 4 } & { 23 } \\ { - 3 } & { 6 } & { 2 } \\ { 1 } & { 5 } & { - 2 } \end{array} \right|$$

Evaluate

$$-543$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&4&23\\-3&6&2\\1&5&-2\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&4&23&2&4\\-3&6&2&-3&6\\1&5&-2&1&5\end{matrix}\right)$$

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

$$2\times 6\left(-2\right)+4\times 2+23\left(-3\right)\times 5=-361$$

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

$$6\times 23+5\times 2\times 2-2\left(-3\right)\times 4=182$$

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

$$-361-182$$

Subtract 182 from -361.

$$-543$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&4&23\\-3&6&2\\1&5&-2\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}6&2\\5&-2\end{matrix}\right))-4det(\left(\begin{matrix}-3&2\\1&-2\end{matrix}\right))+23det(\left(\begin{matrix}-3&6\\1&5\end{matrix}\right))$$

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

$$2\left(6\left(-2\right)-5\times 2\right)-4\left(-3\left(-2\right)-2\right)+23\left(-3\times 5-6\right)$$

Simplify.

$$2\left(-22\right)-4\times 4+23\left(-21\right)$$

Add the terms to obtain the final result.

$$-543$$

Factor

$$-543$$