Solve {ccc}{4}&{6}&{-2}{-4}&{1}&{3}{3}&{-1}&{0}

$$\left| \begin{array} { c c c } { 4 } & { 6 } & { - 2 } \\ { - 4 } & { 1 } & { 3 } \\ { 3 } & { - 1 } & { 0 } \end{array} \right|$$

Evaluate

$$64$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&6&-2\\-4&1&3\\3&-1&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&6&-2&4&6\\-4&1&3&-4&1\\3&-1&0&3&-1\end{matrix}\right)$$

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

$$6\times 3\times 3-2\left(-4\right)\left(-1\right)=46$$

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

$$3\left(-2\right)-3\times 4=-18$$

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

$$46-\left(-18\right)$$

Subtract -18 from 46.

$$64$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&6&-2\\-4&1&3\\3&-1&0\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}1&3\\-1&0\end{matrix}\right))-6det(\left(\begin{matrix}-4&3\\3&0\end{matrix}\right))-2det(\left(\begin{matrix}-4&1\\3&-1\end{matrix}\right))$$

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

$$4\left(-\left(-3\right)\right)-6\left(-3\times 3\right)-2\left(-4\left(-1\right)-3\right)$$

Simplify.

$$4\times 3-6\left(-9\right)-2$$

Add the terms to obtain the final result.

$$64$$

Factor

$$2^{6}$$