Solve {lll}{-3}&{-5}&{-5}{-5}&{-3}&{-5}{-5}&{-5}&{-3}

$$\left| \begin{array} { l l l } { - 3 } & { - 5 } & { - 5 } \\ { - 5 } & { - 3 } & { - 5 } \\ { - 5 } & { - 5 } & { - 3 } \end{array} \right|$$

Evaluate

$$-52$$

Steps Using Sarrus' Rule

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

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

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

$$\left(\begin{matrix}-3&-5&-5&-3&-5\\-5&-3&-5&-5&-3\\-5&-5&-3&-5&-5\end{matrix}\right)$$

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

$$-3\left(-3\right)\left(-3\right)-5\left(-5\right)\left(-5\right)-5\left(-5\right)\left(-5\right)=-277$$

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

$$-5\left(-3\right)\left(-5\right)-5\left(-5\right)\left(-3\right)-3\left(-5\right)\left(-5\right)=-225$$

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

$$-277-\left(-225\right)$$

Subtract -225 from -277.

$$-52$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$-3\left(-3\left(-3\right)-\left(-5\left(-5\right)\right)\right)-\left(-5\left(-5\left(-3\right)-\left(-5\left(-5\right)\right)\right)\right)-5\left(-5\left(-5\right)-\left(-5\left(-3\right)\right)\right)$$

Simplify.

$$-3\left(-16\right)-\left(-5\left(-10\right)\right)-5\times 10$$

Add the terms to obtain the final result.

$$-52$$

Factor

$$-52$$