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

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

Evaluate

$$31$$

Steps Using Sarrus' Rule

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

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

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

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

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

$$5\times 3\times 4-2\times 5=50$$

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

$$3+4\left(-2\right)\left(-2\right)=19$$

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

$$50-19$$

Subtract 19 from 50.

$$31$$

Steps Using Expansion by Minors

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

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

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.

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

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

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

Simplify.

$$5\times 12-\left(-2\left(-8\right)\right)-13$$

Add the terms to obtain the final result.

$$31$$

Factor

$$31$$