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

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

Evaluate

$$1$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\left(-1\right)\times 2-2\left(-3\right)\left(-2\right)-4\times 3=-34$$

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

$$-2\left(-1\right)\left(-4\right)-3\times 5+2\times 3\left(-2\right)=-35$$

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

$$-34-\left(-35\right)$$

Subtract -35 from -34.

$$1$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5-4$$

Add the terms to obtain the final result.

$$1$$

Factor

$$1$$