Solve {ccc}{16}&{200}&{200}{-2}&{0}&{-200}{0}&{-1}&{1}

$$\left| \begin{array} { c c c } { 16 } & { 200 } & { 200 } \\ { - 2 } & { 0 } & { - 200 } \\ { 0 } & { - 1 } & { 1 } \end{array} \right|$$

Evaluate

$$-2400$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}16&200&200\\-2&0&-200\\0&-1&1\end{matrix}\right))$$

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

$$\left(\begin{matrix}16&200&200&16&200\\-2&0&-200&-2&0\\0&-1&1&0&-1\end{matrix}\right)$$

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

$$200\left(-2\right)\left(-1\right)=400$$

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

$$-\left(-200\right)\times 16-2\times 200=2800$$

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

$$400-2800$$

Subtract 2800 from 400.

$$-2400$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}16&200&200\\-2&0&-200\\0&-1&1\end{matrix}\right))$$

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.

$$16det(\left(\begin{matrix}0&-200\\-1&1\end{matrix}\right))-200det(\left(\begin{matrix}-2&-200\\0&1\end{matrix}\right))+200det(\left(\begin{matrix}-2&0\\0&-1\end{matrix}\right))$$

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

$$16\left(-\left(-\left(-200\right)\right)\right)-200\left(-2\right)+200\left(-2\right)\left(-1\right)$$

Simplify.

$$16\left(-200\right)-200\left(-2\right)+200\times 2$$

Add the terms to obtain the final result.

$$-2400$$

Factor

$$-2400$$