Solve {ccc}{-2/3}&{5/3}&{1}{2}&{-5}&{2}{4}&{4}&{1}

$$\left( \begin{array} { c c c } { - \frac { 2 } { 3 } } & { \frac { 5 } { 3 } } & { 1 } \\ { 2 } & { - 5 } & { 2 } \\ { 4 } & { 4 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$\frac{140}{3} = 46\frac{2}{3} = 46.666666666666664$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-\frac{2}{3}&\frac{5}{3}&1\\2&-5&2\\4&4&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-\frac{2}{3}&\frac{5}{3}&1&-\frac{2}{3}&\frac{5}{3}\\2&-5&2&2&-5\\4&4&1&4&4\end{matrix}\right)$$

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

$$-\frac{2}{3}\left(-5\right)+\frac{5}{3}\times 2\times 4+2\times 4=\frac{74}{3}$$

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

$$4\left(-5\right)+4\times 2\left(-\frac{2}{3}\right)+2\times \frac{5}{3}=-22$$

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

$$\frac{74}{3}-\left(-22\right)$$

Subtract -22 from \frac{74}{3}.

$$\frac{140}{3}$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-\frac{2}{3}&\frac{5}{3}&1\\2&-5&2\\4&4&1\end{matrix}\right))$$

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.

$$-\frac{2}{3}det(\left(\begin{matrix}-5&2\\4&1\end{matrix}\right))-\frac{5}{3}det(\left(\begin{matrix}2&2\\4&1\end{matrix}\right))+det(\left(\begin{matrix}2&-5\\4&4\end{matrix}\right))$$

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

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

Simplify.

$$-\frac{2}{3}\left(-13\right)-\frac{5}{3}\left(-6\right)+28$$

Add the terms to obtain the final result.

$$\frac{140}{3}$$

Evaluate

$$\left(\begin{matrix}-\frac{2}{3}&\frac{5}{3}&1\\2&-5&2\\4&4&1\end{matrix}\right)$$