Solve {ccc}{5}&{-2}&{21}{3}&{3}&{0}{3}&{3}&{0}

$$\left( \begin{array} { c c c } { 5 } & { - 2 } & { 21 } \\ { 3 } & { 3 } & { 0 } \\ { 3 } & { 3 } & { 0 } \end{array} \right)$$

Calculate Determinant

$$0$$

Steps Using Sarrus' Rule

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

$$det(\left(\begin{matrix}5&-2&21\\3&3&0\\3&3&0\end{matrix}\right))$$

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

$$\left(\begin{matrix}5&-2&21&5&-2\\3&3&0&3&3\\3&3&0&3&3\end{matrix}\right)$$

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

$$21\times 3\times 3=189$$

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

$$3\times 3\times 21=189$$

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

$$189-189$$

Subtract 189 from 189.

$$0$$

Steps Using Expansion by Minors

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

$$det(\left(\begin{matrix}5&-2&21\\3&3&0\\3&3&0\end{matrix}\right))$$

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.

$$5det(\left(\begin{matrix}3&0\\3&0\end{matrix}\right))-\left(-2det(\left(\begin{matrix}3&0\\3&0\end{matrix}\right))\right)+21det(\left(\begin{matrix}3&3\\3&3\end{matrix}\right))$$

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

$$21\left(3\times 3-3\times 3\right)$$

Simplify.

$$0$$

Evaluate

$$\left(\begin{matrix}5&-2&21\\3&3&0\\3&3&0\end{matrix}\right)$$