Solve {ccc}{-5}&{1}&{7}{1}&{7}&{-5}{7}&{-5}&{1}

$$\left( \begin{array} { c c c } { - 5 } & { 1 } & { 7 } \\ { 1 } & { 7 } & { - 5 } \\ { 7 } & { - 5 } & { 1 } \end{array} \right)$$

Calculate Determinant

$$-324$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-5&1&7\\1&7&-5\\7&-5&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}-5&1&7&-5&1\\1&7&-5&1&7\\7&-5&1&7&-5\end{matrix}\right)$$

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

$$-5\times 7-5\times 7+7\left(-5\right)=-105$$

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

$$7\times 7\times 7-5\left(-5\right)\left(-5\right)+1=219$$

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

$$-105-219$$

Subtract 219 from -105.

$$-324$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-5&1&7\\1&7&-5\\7&-5&1\end{matrix}\right))$$

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.

$$-5det(\left(\begin{matrix}7&-5\\-5&1\end{matrix}\right))-det(\left(\begin{matrix}1&-5\\7&1\end{matrix}\right))+7det(\left(\begin{matrix}1&7\\7&-5\end{matrix}\right))$$

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

$$-5\left(7-\left(-5\left(-5\right)\right)\right)-\left(1-7\left(-5\right)\right)+7\left(-5-7\times 7\right)$$

Simplify.

$$-5\left(-18\right)-36+7\left(-54\right)$$

Add the terms to obtain the final result.

$$-324$$

Evaluate

$$\left(\begin{matrix}-5&1&7\\1&7&-5\\7&-5&1\end{matrix}\right)$$