Solve {ccc}{-2}&{3}&{1}{50}&{3}&{20}{5}&{2}&{398}

$$\left| \begin{array} { c c c } { - 2 } & { 3 } & { 1 } \\ { 50 } & { 3 } & { 20 } \\ { 5 } & { 2 } & { 398 } \end{array} \right|$$

Evaluate

$$-61623$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&3&1\\50&3&20\\5&2&398\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&3&1&-2&3\\50&3&20&50&3\\5&2&398&5&2\end{matrix}\right)$$

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

$$-2\times 3\times 398+3\times 20\times 5+50\times 2=-1988$$

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

$$5\times 3+2\times 20\left(-2\right)+398\times 50\times 3=59635$$

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

$$-1988-59635$$

Subtract 59635 from -1988.

$$-61623$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&3&1\\50&3&20\\5&2&398\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}3&20\\2&398\end{matrix}\right))-3det(\left(\begin{matrix}50&20\\5&398\end{matrix}\right))+det(\left(\begin{matrix}50&3\\5&2\end{matrix}\right))$$

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

$$-2\left(3\times 398-2\times 20\right)-3\left(50\times 398-5\times 20\right)+50\times 2-5\times 3$$

Simplify.

$$-2\times 1154-3\times 19800+85$$

Add the terms to obtain the final result.

$$-61623$$

Factor

$$-61623$$