Solve {lll}{0}&{c}&{0}{c}&{0}&{0}{0}&{0}&{c}?

$$\left( \begin{array} { l l l } { 0 } & { c } & { 0 } \\ { c } & { 0 } & { 0 } \\ { 0 } & { 0 } & { c } \end{array} \right) ?$$

Calculate Determinant

$$-c^{3}$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}0&c&0\\c&0&0\\0&0&c\end{matrix}\right))$$

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

$$\left(\begin{matrix}0&c&0&0&c\\c&0&0&c&0\\0&0&c&0&0\end{matrix}\right)$$

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.

$$ccc=c^{3}$$

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

$$-c^{3}$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}0&c&0\\c&0&0\\0&0&c\end{matrix}\right))$$

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.

$$-cdet(\left(\begin{matrix}c&0\\0&c\end{matrix}\right))$$

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

$$-ccc$$

Simplify.

$$-cc^{2}$$

Add the terms to obtain the final result.

$$-c^{3}$$

Evaluate

$$\left(\begin{matrix}0&c&0\\c&0&0\\0&0&c\end{matrix}\right)$$