Solve {lll}{3}&{4}&{5}{1}&{3}&{4}{7}&{4}&{1}

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

Evaluate

$$-16$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3+4\times 4\times 7+5\times 4=141$$

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

$$7\times 3\times 5+4\times 4\times 3+4=157$$

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

$$141-157$$

Subtract 157 from 141.

$$-16$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$3\left(-13\right)-4\left(-27\right)+5\left(-17\right)$$

Add the terms to obtain the final result.

$$-16$$

Factor

$$-16$$