Solve {ccc}{1}&{3}&{4}{5}&{0}&{3}{1}&{-2}&{1}

$$\left| \begin{array} { c c c } { 1 } & { 3 } & { 4 } \\ { 5 } & { 0 } & { 3 } \\ { 1 } & { - 2 } & { 1 } \end{array} \right|$$

Evaluate

$$-40$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&4\\5&0&3\\1&-2&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&4&1&3\\5&0&3&5&0\\1&-2&1&1&-2\end{matrix}\right)$$

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

$$3\times 3+4\times 5\left(-2\right)=-31$$

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

$$-2\times 3+5\times 3=9$$

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

$$-31-9$$

Subtract 9 from -31.

$$-40$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&4\\5&0&3\\1&-2&1\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}0&3\\-2&1\end{matrix}\right))-3det(\left(\begin{matrix}5&3\\1&1\end{matrix}\right))+4det(\left(\begin{matrix}5&0\\1&-2\end{matrix}\right))$$

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

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

Simplify.

$$6-3\times 2+4\left(-10\right)$$

Add the terms to obtain the final result.

$$-40$$

Factor

$$-40$$