Solve {rrr}{3}&{1}&{7}{7}&{4}&{7}{8}&{3}&{-1}

$$\left| \begin{array} { r r r } { 3 } & { 1 } & { 7 } \\ { 7 } & { 4 } & { 7 } \\ { 8 } & { 3 } & { - 1 } \end{array} \right|$$

Evaluate

$$-89$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&1&7\\7&4&7\\8&3&-1\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&1&7&3&1\\7&4&7&7&4\\8&3&-1&8&3\end{matrix}\right)$$

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

$$3\times 4\left(-1\right)+7\times 8+7\times 7\times 3=191$$

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

$$8\times 4\times 7+3\times 7\times 3-7=280$$

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

$$191-280$$

Subtract 280 from 191.

$$-89$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&1&7\\7&4&7\\8&3&-1\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}4&7\\3&-1\end{matrix}\right))-det(\left(\begin{matrix}7&7\\8&-1\end{matrix}\right))+7det(\left(\begin{matrix}7&4\\8&3\end{matrix}\right))$$

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

$$3\left(4\left(-1\right)-3\times 7\right)-\left(7\left(-1\right)-8\times 7\right)+7\left(7\times 3-8\times 4\right)$$

Simplify.

$$3\left(-25\right)-\left(-63\right)+7\left(-11\right)$$

Add the terms to obtain the final result.

$$-89$$

Factor

$$-89$$