Solve {ccc}{1}&{3}&{5}{1}&{1}&{1}{3}&{-1}&{4}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 5 } \\ { 1 } & { 1 } & { 1 } \\ { 3 } & { - 1 } & { 4 } \end{array} \right|$$

Evaluate

$$-18$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$3\times 5-1+4\times 3=26$$

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

$$8-26$$

Subtract 26 from 8.

$$-18$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-18$$

Factor

$$-18$$