Solve {ccc}{1}&{1}&{1}{-2}&{-3}&{-1}{-1}&{-1}&{-4}

$$\left| \begin{array} { c c c } { 1 } & { 1 } & { 1 } \\ { - 2 } & { - 3 } & { - 1 } \\ { - 1 } & { - 1 } & { - 4 } \end{array} \right|$$

Evaluate

$$3$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&1&1\\-2&-3&-1\\-1&-1&-4\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&1&1&1&1\\-2&-3&-1&-2&-3\\-1&-1&-4&-1&-1\end{matrix}\right)$$

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

$$-3\left(-4\right)-\left(-1\right)-2\left(-1\right)=15$$

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

$$-\left(-3\right)-\left(-1\right)-4\left(-2\right)=12$$

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

$$15-12$$

Subtract 12 from 15.

$$3$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&1&1\\-2&-3&-1\\-1&-1&-4\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}-3&-1\\-1&-4\end{matrix}\right))-det(\left(\begin{matrix}-2&-1\\-1&-4\end{matrix}\right))+det(\left(\begin{matrix}-2&-3\\-1&-1\end{matrix}\right))$$

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

$$-3\left(-4\right)-\left(-\left(-1\right)\right)-\left(-2\left(-4\right)-\left(-\left(-1\right)\right)\right)+-2\left(-1\right)-\left(-\left(-3\right)\right)$$

Simplify.

$$11-7-1$$

Add the terms to obtain the final result.

$$3$$

Factor

$$3$$