Solve {ccc}{-2}&{3}&{1}{503}&{201}&{298}{5}&{2}&{3}

$$\left| \begin{array} { c c c } { - 2 } & { 3 } & { 1 } \\ { 503 } & { 201 } & { 298 } \\ { 5 } & { 2 } & { 3 } \end{array} \right|$$

Evaluate

$$-70$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-2&3&1\\503&201&298\\5&2&3\end{matrix}\right))$$

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

$$\left(\begin{matrix}-2&3&1&-2&3\\503&201&298&503&201\\5&2&3&5&2\end{matrix}\right)$$

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

$$-2\times 201\times 3+3\times 298\times 5+503\times 2=4270$$

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

$$5\times 201+2\times 298\left(-2\right)+3\times 503\times 3=4340$$

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

$$4270-4340$$

Subtract 4340 from 4270.

$$-70$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-2&3&1\\503&201&298\\5&2&3\end{matrix}\right))$$

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.

$$-2det(\left(\begin{matrix}201&298\\2&3\end{matrix}\right))-3det(\left(\begin{matrix}503&298\\5&3\end{matrix}\right))+det(\left(\begin{matrix}503&201\\5&2\end{matrix}\right))$$

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

$$-2\left(201\times 3-2\times 298\right)-3\left(503\times 3-5\times 298\right)+503\times 2-5\times 201$$

Simplify.

$$-2\times 7-3\times 19+1$$

Add the terms to obtain the final result.

$$-70$$

Factor

$$-70$$