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

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

Evaluate

$$15$$

Steps Using Sarrus' Rule

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

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

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

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

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

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

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

$$-2-2+4\left(-3\right)=-16$$

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

$$-1-\left(-16\right)$$

Subtract -16 from -1.

$$15$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-6-\left(-13\right)+8$$

Add the terms to obtain the final result.

$$15$$

Factor

$$3\times 5$$