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

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

Evaluate

$$-3$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3+4\times 8+5\times 3=50$$

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

$$8\times 5+3\times 3+4=53$$

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

$$50-53$$

Subtract 53 from 50.

$$-3$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$3\left(1-3\right)-4\left(1-8\right)+5\left(3-8\right)$$

Simplify.

$$3\left(-2\right)-4\left(-7\right)+5\left(-5\right)$$

Add the terms to obtain the final result.

$$-3$$

Factor

$$-3$$