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

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

Evaluate

$$-26$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-4\times 5+1+1=-18$$

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

$$4-1+5=8$$

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

$$-18-8$$

Subtract 8 from -18.

$$-26$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-19-4-3$$

Add the terms to obtain the final result.

$$-26$$

Factor

$$-26$$