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

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

Calculate Determinant

$$\frac{10}{3} = 3\frac{1}{3} = 3.3333333333333335$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$-\frac{2}{3}\left(-\frac{2}{3}\right)-\frac{2}{3}+\frac{4}{3}=\frac{10}{9}$$

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

$$-\frac{2}{3}\times \frac{4}{3}-\frac{2}{3}-\frac{2}{3}=-\frac{20}{9}$$

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

$$\frac{10}{9}-\left(-\frac{20}{9}\right)$$

Subtract -\frac{20}{9} from \frac{10}{9} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{10}{3}$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$-\frac{2}{3}\left(-\frac{5}{3}\right)+\frac{4}{3}\times \frac{5}{3}$$

Add the terms to obtain the final result.

$$\frac{10}{3}$$

Evaluate

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