Solve {lll}{3}&{4}&{5}{7}&{8}&{9}{7}&{1}&{41}

$$\left| \begin{array} { l l l } { 3 } & { 4 } & { 5 } \\ { 7 } & { 8 } & { 9 } \\ { 7 } & { 1 } & { 41 } \end{array} \right|$$

Evaluate

$$-184$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}3&4&5\\7&8&9\\7&1&41\end{matrix}\right))$$

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

$$\left(\begin{matrix}3&4&5&3&4\\7&8&9&7&8\\7&1&41&7&1\end{matrix}\right)$$

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

$$3\times 8\times 41+4\times 9\times 7+5\times 7=1271$$

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

$$7\times 8\times 5+9\times 3+41\times 7\times 4=1455$$

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

$$1271-1455$$

Subtract 1455 from 1271.

$$-184$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}3&4&5\\7&8&9\\7&1&41\end{matrix}\right))$$

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.

$$3det(\left(\begin{matrix}8&9\\1&41\end{matrix}\right))-4det(\left(\begin{matrix}7&9\\7&41\end{matrix}\right))+5det(\left(\begin{matrix}7&8\\7&1\end{matrix}\right))$$

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

$$3\left(8\times 41-9\right)-4\left(7\times 41-7\times 9\right)+5\left(7-7\times 8\right)$$

Simplify.

$$3\times 319-4\times 224+5\left(-49\right)$$

Add the terms to obtain the final result.

$$-184$$

Factor

$$-184$$