Solve {lll}{1/2}&{1}&{1}{1}&{1/2}&{1}{1}&{1}&{1/2}

$$\left| \begin{array} { l l l } { \frac { 1 } { 2 } } & { 1 } & { 1 } \\ { 1 } & { \frac { 1 } { 2 } } & { 1 } \\ { 1 } & { 1 } & { \frac { 1 } { 2 } } \end{array} \right|$$

Evaluate

$$\frac{5}{8}=0.625$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+1+1=\frac{17}{8}$$

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

$$\frac{1+1+1}{2}=\frac{3}{2}$$

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

$$\frac{17}{8}-\frac{3}{2}$$

Subtract \frac{3}{2} from \frac{17}{8} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{5}{8}$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$\frac{5}{8}$$

Factor

$$\frac{5}{2 ^ {3}} = 0.625$$