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

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

Evaluate

$$49$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-3\times 5+2\left(-2\right)-3\times 2\times 4=-43$$

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

$$6-\left(-43\right)$$

Subtract -43 from 6.

$$49$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-2\left(-3\left(-3\right)-2\right)-4\left(2\left(-3\right)-1\right)+5\left(2\times 2-\left(-3\right)\right)$$

Simplify.

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

Add the terms to obtain the final result.

$$49$$

Factor

$$7^{2}$$