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

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

Evaluate

$$-30$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 2+4\times 3\times 2+4=38$$

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

$$2\times 2+4\times 3\times 5+4=68$$

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

$$38-68$$

Subtract 68 from 38.

$$-30$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

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

Add the terms to obtain the final result.

$$-30$$

Factor

$$-30$$