Solve {rrr}{2}&{-1}&{3}{1}&{1}&{1}{3}&{-8}&{7}

$$\left| \begin{array} { r r r } { 2 } & { - 1 } & { 3 } \\ { 1 } & { 1 } & { 1 } \\ { 3 } & { - 8 } & { 7 } \end{array} \right|$$

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}2&-1&3\\1&1&1\\3&-8&7\end{matrix}\right))$$

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

$$\left(\begin{matrix}2&-1&3&2&-1\\1&1&1&1&1\\3&-8&7&3&-8\end{matrix}\right)$$

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

$$2\times 7-3+3\left(-8\right)=-13$$

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

$$3\times 3-8\times 2+7\left(-1\right)=-14$$

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

$$-13-\left(-14\right)$$

Subtract -14 from -13.

$$1$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}2&-1&3\\1&1&1\\3&-8&7\end{matrix}\right))$$

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.

$$2det(\left(\begin{matrix}1&1\\-8&7\end{matrix}\right))-\left(-det(\left(\begin{matrix}1&1\\3&7\end{matrix}\right))\right)+3det(\left(\begin{matrix}1&1\\3&-8\end{matrix}\right))$$

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

$$2\left(7-\left(-8\right)\right)-\left(-\left(7-3\right)\right)+3\left(-8-3\right)$$

Simplify.

$$2\times 15-\left(-4\right)+3\left(-11\right)$$

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$