Solve {ccc}{13}&{5}&{-7}{6}&{1}&{-12}{20}&{9}&{-3}

$$\left| \begin{array} { c c c } { 13 } & { 5 } & { - 7 } \\ { 6 } & { 1 } & { - 12 } \\ { 20 } & { 9 } & { - 3 } \end{array} \right|$$

Evaluate

$$17$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}13&5&-7\\6&1&-12\\20&9&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}13&5&-7&13&5\\6&1&-12&6&1\\20&9&-3&20&9\end{matrix}\right)$$

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

$$13\left(-3\right)+5\left(-12\right)\times 20-7\times 6\times 9=-1617$$

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

$$20\left(-7\right)+9\left(-12\right)\times 13-3\times 6\times 5=-1634$$

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

$$-1617-\left(-1634\right)$$

Subtract -1634 from -1617.

$$17$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}13&5&-7\\6&1&-12\\20&9&-3\end{matrix}\right))$$

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.

$$13det(\left(\begin{matrix}1&-12\\9&-3\end{matrix}\right))-5det(\left(\begin{matrix}6&-12\\20&-3\end{matrix}\right))-7det(\left(\begin{matrix}6&1\\20&9\end{matrix}\right))$$

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

$$13\left(-3-9\left(-12\right)\right)-5\left(6\left(-3\right)-20\left(-12\right)\right)-7\left(6\times 9-20\right)$$

Simplify.

$$13\times 105-5\times 222-7\times 34$$

Add the terms to obtain the final result.

$$17$$

Factor

$$17$$