Solve {rrr}{4}&{1}&{5}{3}&{2}&{-6}{12}&{3}&{2}

$$\left| \begin{array} { r r r } { 4 } & { 1 } & { 5 } \\ { 3 } & { 2 } & { - 6 } \\ { 12 } & { 3 } & { 2 } \end{array} \right|$$

Evaluate

$$-65$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}4&1&5\\3&2&-6\\12&3&2\end{matrix}\right))$$

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

$$\left(\begin{matrix}4&1&5&4&1\\3&2&-6&3&2\\12&3&2&12&3\end{matrix}\right)$$

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

$$4\times 2\times 2-6\times 12+5\times 3\times 3=-11$$

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

$$12\times 2\times 5+3\left(-6\right)\times 4+2\times 3=54$$

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

$$-11-54$$

Subtract 54 from -11.

$$-65$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}4&1&5\\3&2&-6\\12&3&2\end{matrix}\right))$$

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.

$$4det(\left(\begin{matrix}2&-6\\3&2\end{matrix}\right))-det(\left(\begin{matrix}3&-6\\12&2\end{matrix}\right))+5det(\left(\begin{matrix}3&2\\12&3\end{matrix}\right))$$

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

$$4\left(2\times 2-3\left(-6\right)\right)-\left(3\times 2-12\left(-6\right)\right)+5\left(3\times 3-12\times 2\right)$$

Simplify.

$$4\times 22-78+5\left(-15\right)$$

Add the terms to obtain the final result.

$$-65$$

Factor

$$-65$$