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

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

Evaluate

$$-52$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-1\right)\times 4+9\times 9\times 3-3\times 6=209$$

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

$$3\left(-1\right)\left(-3\right)+9\times 4+4\times 6\times 9=261$$

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

$$209-261$$

Subtract 261 from 209.

$$-52$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$4\left(-4-9\right)-9\left(6\times 4-3\times 9\right)-3\left(6-3\left(-1\right)\right)$$

Simplify.

$$4\left(-13\right)-9\left(-3\right)-3\times 9$$

Add the terms to obtain the final result.

$$-52$$

Factor

$$-52$$