Solve {rrr}{1}&{3}&{302}{-4}&{3}&{297}{2}&{2}&{203}

$$\left| \begin{array} { r r r } { 1 } & { 3 } & { 302 } \\ { - 4 } & { 3 } & { 297 } \\ { 2 } & { 2 } & { 203 } \end{array} \right|$$

Evaluate

$$5$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}1&3&302\\-4&3&297\\2&2&203\end{matrix}\right))$$

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

$$\left(\begin{matrix}1&3&302&1&3\\-4&3&297&-4&3\\2&2&203&2&2\end{matrix}\right)$$

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

$$3\times 203+3\times 297\times 2+302\left(-4\right)\times 2=-25$$

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

$$2\times 3\times 302+2\times 297+203\left(-4\right)\times 3=-30$$

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

$$-25-\left(-30\right)$$

Subtract -30 from -25.

$$5$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}1&3&302\\-4&3&297\\2&2&203\end{matrix}\right))$$

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.

$$det(\left(\begin{matrix}3&297\\2&203\end{matrix}\right))-3det(\left(\begin{matrix}-4&297\\2&203\end{matrix}\right))+302det(\left(\begin{matrix}-4&3\\2&2\end{matrix}\right))$$

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

$$3\times 203-2\times 297-3\left(-4\times 203-2\times 297\right)+302\left(-4\times 2-2\times 3\right)$$

Simplify.

$$15-3\left(-1406\right)+302\left(-14\right)$$

Add the terms to obtain the final result.

$$5$$

Factor

$$5$$