Solve {ccc}{-1/3}&{2/3}&{0}{0}&{0}&{1/3}{2/3}&{-1/3}&{0}

$$\left( \begin{array} { c c c } { - 1 / 3 } & { 2 / 3 } & { 0 } \\ { 0 } & { 0 } & { 1 / 3 } \\ { 2 / 3 } & { - 1 / 3 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$\frac{1}{9} = 0.1111111111111111$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}&0\\0&0&\frac{1}{3}\\\frac{2}{3}&-\frac{1}{3}&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}&0&-\frac{1}{3}&\frac{2}{3}\\0&0&\frac{1}{3}&0&0\\\frac{2}{3}&-\frac{1}{3}&0&\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)$$

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

$$\frac{2}{3}\times \frac{1}{3}\times \frac{2}{3}=\frac{4}{27}$$

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

$$-\frac{1}{3}\times \frac{1}{3}\left(-\frac{1}{3}\right)=\frac{1}{27}$$

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

$$\frac{4}{27}-\frac{1}{27}$$

Subtract \frac{1}{27} from \frac{4}{27} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{9}$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}&0\\0&0&\frac{1}{3}\\\frac{2}{3}&-\frac{1}{3}&0\end{matrix}\right))$$

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.

$$-\frac{1}{3}det(\left(\begin{matrix}0&\frac{1}{3}\\-\frac{1}{3}&0\end{matrix}\right))-\frac{2}{3}det(\left(\begin{matrix}0&\frac{1}{3}\\\frac{2}{3}&0\end{matrix}\right))$$

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

$$-\frac{1}{3}\left(-\left(-\frac{1}{3}\times \frac{1}{3}\right)\right)-\frac{2}{3}\left(-\frac{2}{3}\times \frac{1}{3}\right)$$

Simplify.

$$-\frac{1}{3}\times \frac{1}{9}-\frac{2}{3}\left(-\frac{2}{9}\right)$$

Add the terms to obtain the final result.

$$\frac{1}{9}$$

Evaluate

$$\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}&0\\0&0&\frac{1}{3}\\\frac{2}{3}&-\frac{1}{3}&0\end{matrix}\right)$$