Solve {cc}{20}&{4}{2}&{2}&{2}{0}&{-7}&{0}

$$\left| \begin{array} { c c } { 20 } & { 4 } \\ { 2 } & { 2 } & { 2 } \\ { 0 } & { - 7 } & { 0 } \end{array} \right|$$

Evaluate

$$280$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}20&4&0\\2&2&2\\0&-7&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}20&4&0&20&4\\2&2&2&2&2\\0&-7&0&0&-7\end{matrix}\right)$$

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

$$\text{true}$$

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

$$-7\times 2\times 20=-280$$

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

$$-\left(-280\right)$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}20&4&0\\2&2&2\\0&-7&0\end{matrix}\right))$$

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.

$$20det(\left(\begin{matrix}2&2\\-7&0\end{matrix}\right))-4det(\left(\begin{matrix}2&2\\0&0\end{matrix}\right))$$

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

$$20\left(-\left(-7\times 2\right)\right)$$

Simplify.

$$20\times 14$$

Add the terms to obtain the final result.

$$280$$

Factor

$$2^{3}\times 5\times 7$$