Solve {cc}{-3}&{00}{0}&{30}{0}&{0}&{-3}

$$\left| \begin{array} { c c } { - 3 } & { 00 } \\ { 0 } & { 30 } \\ { 0 } & { 0 } & { - 3 } \end{array} \right|$$

Evaluate

$$270$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}-3&0&0\\0&30&0\\0&0&-3\end{matrix}\right))$$

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

$$\left(\begin{matrix}-3&0&0&-3&0\\0&30&0&0&30\\0&0&-3&0&0\end{matrix}\right)$$

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

$$-3\times 30\left(-3\right)=270$$

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

$$\text{true}$$

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

$$270$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}-3&0&0\\0&30&0\\0&0&-3\end{matrix}\right))$$

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.

$$-3det(\left(\begin{matrix}30&0\\0&-3\end{matrix}\right))$$

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

$$-3\times 30\left(-3\right)$$

Simplify.

$$-3\left(-90\right)$$

Add the terms to obtain the final result.

$$270$$

Factor

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