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

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

Evaluate

$$219$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\left(-3\right)\times 3+6\times 4\times 6=153$$

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

$$4\left(-3\right)\times 6+6\left(-1\right)\left(-1\right)=-66$$

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

$$153-\left(-66\right)$$

Subtract -66 from 153.

$$219$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\left(-3\right)+6\times 36$$

Add the terms to obtain the final result.

$$219$$

Factor

$$3\times 73$$