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

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

Evaluate

$$60$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$3\times 3\times 3-\left(-2\times 5\right)-4\left(-2\right)=45$$

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

$$5\times 3\left(-1\right)-2\left(-2\right)\times 3+3\times 4\left(-1\right)=-15$$

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

$$45-\left(-15\right)$$

Subtract -15 from 45.

$$60$$

Steps Using Expansion by Minors

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

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

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.

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

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

$$3\left(3\times 3-\left(-2\left(-2\right)\right)\right)-\left(-\left(4\times 3-5\left(-2\right)\right)\right)-\left(4\left(-2\right)-5\times 3\right)$$

Simplify.

$$3\times 5-\left(-22\right)-\left(-23\right)$$

Add the terms to obtain the final result.

$$60$$

Factor

$$2^{2}\times 3\times 5$$