Solve {rrr}{-9}&{3}&{-4}{7}&{-5}&{-3}{4}&{6}&{1}

$$\left| \begin{array} { r r r } { - 9 } & { 3 } & { - 4 } \\ { 7 } & { - 5 } & { - 3 } \\ { 4 } & { 6 } & { 1 } \end{array} \right|$$

Evaluate

$$-422$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-9&3&-4\\7&-5&-3\\4&6&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-9&3&-4&-9&3\\7&-5&-3&7&-5\\4&6&1&4&6\end{matrix}\right)$$

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

$$-9\left(-5\right)+3\left(-3\right)\times 4-4\times 7\times 6=-159$$

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

$$4\left(-5\right)\left(-4\right)+6\left(-3\right)\left(-9\right)+7\times 3=263$$

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

$$-159-263$$

Subtract 263 from -159.

$$-422$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-9&3&-4\\7&-5&-3\\4&6&1\end{matrix}\right))$$

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.

$$-9det(\left(\begin{matrix}-5&-3\\6&1\end{matrix}\right))-3det(\left(\begin{matrix}7&-3\\4&1\end{matrix}\right))-4det(\left(\begin{matrix}7&-5\\4&6\end{matrix}\right))$$

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

$$-9\left(-5-6\left(-3\right)\right)-3\left(7-4\left(-3\right)\right)-4\left(7\times 6-4\left(-5\right)\right)$$

Simplify.

$$-9\times 13-3\times 19-4\times 62$$

Add the terms to obtain the final result.

$$-422$$

Factor

$$-422$$