In Exercises 21–23, use determinants to find out if the matrix is invertible.

21. \(\left( {\begin{aligned}{*{20}{c}}2&6&0\\1&3&2\\3&9&2\end{aligned}} \right)\)

Matrix A of the order\(n \times n\)(square matrix) isinvertible if thedeterminant of the matrix is not 0 (\(\det \left( A \right) \ne 0\)).

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}2&6&0\\1&3&2\\3&9&2\end{aligned}} \right)\).

Compute the determinate of the matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = \left| {\begin{aligned}{*{20}{c}}2&6&0\\1&3&2\\3&9&2\end{aligned}} \right|\\ = 2 \cdot \left| {\begin{aligned}{*{20}{c}}3&2\\9&2\end{aligned}} \right| - 6 \cdot \left| {\begin{aligned}{*{20}{c}}1&2\\3&2\end{aligned}} \right| + 0 \cdot \left| {\begin{aligned}{*{20}{c}}1&3\\3&9\end{aligned}} \right|\\ = 2 \cdot \left( {2\left( 3 \right) - 9\left( 2 \right)} \right) - 6\left( {1\left( 2 \right) - 3\left( 2 \right)} \right) + 0\\ = 2\left( { - 12} \right) - 6\left( { - 4} \right)\\ = - 24 + 24\\ = 0\end{aligned}\)

Since \(\det \left( A \right) = 0\), the matrix is not invertible.