Solve {ccc}{-2}&{-3}&{-1}{1}&{1}&{1}{3}&{-7}&{5}

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

Evaluate

$$-8$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-2\times 5-3\times 3-\left(-7\right)=-12$$

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

$$3\left(-1\right)-7\left(-2\right)+5\left(-3\right)=-4$$

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

$$-12-\left(-4\right)$$

Subtract -4 from -12.

$$-8$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-2\left(5-\left(-7\right)\right)-\left(-3\left(5-3\right)\right)-\left(-7-3\right)$$

Simplify.

$$-2\times 12-\left(-3\times 2\right)-\left(-10\right)$$

Add the terms to obtain the final result.

$$-8$$

Factor

$$-8$$