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

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

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$2\left(-2\right)-3\left(-3\right)+1=6$$

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

$$-3+2-2\left(-3\right)=5$$

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

$$6-5$$

Subtract 5 from 6.

$$1$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$