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

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

Evaluate

$$7$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$4\left(-3\right)\left(-2\right)+3\times 3\left(-5\right)-5=-26$$

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

$$-5\left(-3\right)\left(-1\right)+3\times 4-2\times 5\times 3=-33$$

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

$$-26-\left(-33\right)$$

Subtract -33 from -26.

$$7$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$4\times 3-3\times 5-\left(-10\right)$$

Add the terms to obtain the final result.

$$7$$

Factor

$$7$$