Solve {ccc}{1}&{12}&{-1}{3}&{-13}&{4}{-1}&{5}&{1}

$$\left| \begin{array} { c c c } { 1 } & { 12 } & { - 1 } \\ { 3 } & { - 13 } & { 4 } \\ { - 1 } & { 5 } & { 1 } \end{array} \right|$$

Evaluate

$$-119$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&12&-1\\3&-13&4\\-1&5&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&12&-1&1&12\\3&-13&4&3&-13\\-1&5&1&-1&5\end{matrix}\right)$$

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

$$-13+12\times 4\left(-1\right)-3\times 5=-76$$

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

$$-\left(-13\right)\left(-1\right)+5\times 4+3\times 12=43$$

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

$$-76-43$$

Subtract 43 from -76.

$$-119$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&12&-1\\3&-13&4\\-1&5&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-13&4\\5&1\end{matrix}\right))-12det(\left(\begin{matrix}3&4\\-1&1\end{matrix}\right))-det(\left(\begin{matrix}3&-13\\-1&5\end{matrix}\right))$$

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

$$-13-5\times 4-12\left(3-\left(-4\right)\right)-\left(3\times 5-\left(-\left(-13\right)\right)\right)$$

Simplify.

$$-33-12\times 7-2$$

Add the terms to obtain the final result.

$$-119$$

Factor

$$-119$$