Solve {lll}{1}&{1}&{1}{3}&{1}&{4}{8}&{9}&{5}=

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

Evaluate

$$5$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5+4\times 8+3\times 9=64$$

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

$$8+9\times 4+5\times 3=59$$

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

$$64-59$$

Subtract 59 from 64.

$$5$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-31-\left(-17\right)+19$$

Add the terms to obtain the final result.

$$5$$

Factor

$$5$$