Solve {ccc}{4}&{1}&{-6}{1}&{-3}&{5}{0}&{3}&{6}

$$\left| \begin{array} { c c c } { 4 } & { 1 } & { - 6 } \\ { 1 } & { - 3 } & { 5 } \\ { 0 } & { 3 } & { 6 } \end{array} \right|$$

Evaluate

$$-156$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&1&-6\\1&-3&5\\0&3&6\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&1&-6&4&1\\1&-3&5&1&-3\\0&3&6&0&3\end{matrix}\right)$$

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

$$4\left(-3\right)\times 6-6\times 3=-90$$

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

$$3\times 5\times 4+6=66$$

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

$$-90-66$$

Subtract 66 from -90.

$$-156$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&1&-6\\1&-3&5\\0&3&6\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}-3&5\\3&6\end{matrix}\right))-det(\left(\begin{matrix}1&5\\0&6\end{matrix}\right))-6det(\left(\begin{matrix}1&-3\\0&3\end{matrix}\right))$$

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

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

Simplify.

$$4\left(-33\right)-6-6\times 3$$

Add the terms to obtain the final result.

$$-156$$

Factor

$$-156$$