Solve {ccc}{1}&{3}&{7}{6}&{9}&{10}{-1}&{-2}&{-5}=

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 7 } \\ { 6 } & { 9 } & { 10 } \\ { - 1 } & { - 2 } & { - 5 } \end{array} \right| =$$

Evaluate

$$14$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&7\\6&9&10\\-1&-2&-5\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&7&1&3\\6&9&10&6&9\\-1&-2&-5&-1&-2\end{matrix}\right)$$

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

$$9\left(-5\right)+3\times 10\left(-1\right)+7\times 6\left(-2\right)=-159$$

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

$$-9\times 7-2\times 10-5\times 6\times 3=-173$$

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

$$-159-\left(-173\right)$$

Subtract -173 from -159.

$$14$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&7\\6&9&10\\-1&-2&-5\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}9&10\\-2&-5\end{matrix}\right))-3det(\left(\begin{matrix}6&10\\-1&-5\end{matrix}\right))+7det(\left(\begin{matrix}6&9\\-1&-2\end{matrix}\right))$$

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

$$9\left(-5\right)-\left(-2\times 10\right)-3\left(6\left(-5\right)-\left(-10\right)\right)+7\left(6\left(-2\right)-\left(-9\right)\right)$$

Simplify.

$$-25-3\left(-20\right)+7\left(-3\right)$$

Add the terms to obtain the final result.

$$14$$

Factor

$$2\times 7$$