Solve {ccc}{-6}&{4}&{1}{-13}&{9}&{3}{-16}&{9}&{3}

$$\left| \begin{array} { c c c } { - 6 } & { 4 } & { 1 } \\ { - 13 } & { 9 } & { 3 } \\ { - 16 } & { 9 } & { 3 } \end{array} \right|$$

Evaluate

$$-9$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-6&4&1\\-13&9&3\\-16&9&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}-6&4&1&-6&4\\-13&9&3&-13&9\\-16&9&3&-16&9\end{matrix}\right)$$

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

$$-6\times 9\times 3+4\times 3\left(-16\right)-13\times 9=-471$$

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

$$-16\times 9+9\times 3\left(-6\right)+3\left(-13\right)\times 4=-462$$

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

$$-471-\left(-462\right)$$

Subtract -462 from -471.

$$-9$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-6&4&1\\-13&9&3\\-16&9&3\end{matrix}\right))$$

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.

$$-6det(\left(\begin{matrix}9&3\\9&3\end{matrix}\right))-4det(\left(\begin{matrix}-13&3\\-16&3\end{matrix}\right))+det(\left(\begin{matrix}-13&9\\-16&9\end{matrix}\right))$$

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

$$-6\left(9\times 3-9\times 3\right)-4\left(-13\times 3-\left(-16\times 3\right)\right)+-13\times 9-\left(-16\times 9\right)$$

Simplify.

$$-4\times 9+27$$

Add the terms to obtain the final result.

$$-9$$

Factor

$$-9$$