Solve {ccc}{19/22}&{5/11}&{-2/11}{2/11}&{frac{8}{11}}&{-frac{1}{11}}{0}&{-frac{1}{2}}&{1}

$$\left( \begin{array} { c c c } { \frac { 19 } { 22 } } & { \frac { 5 } { 11 } } & { - \frac { 2 } { 11 } } \\ { \frac { 2 } { 11 } } & { \frac { 8 } { 11 } } & { - \frac { 1 } { 11 } } \\ { 0 } & { - \frac { 1 } { 2 } } & { 1 } \end{array} \right)$$

Calculate Determinant

$$\frac{23}{44} = 0.5227272727272727$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}\frac{19}{22}&\frac{5}{11}&-\frac{2}{11}\\\frac{2}{11}&\frac{8}{11}&-\frac{1}{11}\\0&-\frac{1}{2}&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}\frac{19}{22}&\frac{5}{11}&-\frac{2}{11}&\frac{19}{22}&\frac{5}{11}\\\frac{2}{11}&\frac{8}{11}&-\frac{1}{11}&\frac{2}{11}&\frac{8}{11}\\0&-\frac{1}{2}&1&0&-\frac{1}{2}\end{matrix}\right)$$

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

$$\frac{19}{22}\times \frac{8}{11}-\frac{2}{11}\times \frac{2}{11}\left(-\frac{1}{2}\right)=\frac{78}{121}$$

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

$$-\frac{1}{2}\left(-\frac{1}{11}\right)\times \frac{19}{22}+\frac{2}{11}\times \frac{5}{11}=\frac{59}{484}$$

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

$$\frac{78}{121}-\frac{59}{484}$$

Subtract \frac{59}{484} from \frac{78}{121} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{23}{44}$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}\frac{19}{22}&\frac{5}{11}&-\frac{2}{11}\\\frac{2}{11}&\frac{8}{11}&-\frac{1}{11}\\0&-\frac{1}{2}&1\end{matrix}\right))$$

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.

$$\frac{19}{22}det(\left(\begin{matrix}\frac{8}{11}&-\frac{1}{11}\\-\frac{1}{2}&1\end{matrix}\right))-\frac{5}{11}det(\left(\begin{matrix}\frac{2}{11}&-\frac{1}{11}\\0&1\end{matrix}\right))-\frac{2}{11}det(\left(\begin{matrix}\frac{2}{11}&\frac{8}{11}\\0&-\frac{1}{2}\end{matrix}\right))$$

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

$$\frac{19}{22}\left(\frac{8}{11}-\left(-\frac{1}{2}\left(-\frac{1}{11}\right)\right)\right)-\frac{5}{11}\times \frac{2}{11}-\frac{2}{11}\times \frac{2}{11}\left(-\frac{1}{2}\right)$$

Simplify.

$$\frac{19}{22}\times \frac{15}{22}-\frac{5}{11}\times \frac{2}{11}-\frac{2}{11}\left(-\frac{1}{11}\right)$$

Add the terms to obtain the final result.

$$\frac{23}{44}$$

Evaluate

$$\left(\begin{matrix}\frac{19}{22}&\frac{5}{11}&-\frac{2}{11}\\\frac{2}{11}&\frac{8}{11}&-\frac{1}{11}\\0&-\frac{1}{2}&1\end{matrix}\right)$$