If \[A = \left[ {\begin{array}{*{20}{c}}0&3\\2&0\end{array}} \right]\] , and \[{A^{ - 1}} = \lambda \left( {adj\left( A \right)} \right)\] then find the value of \[\lambda \].
A. \[\dfrac{{ - 1}}{6}\]
B. \[\dfrac{1}{3}\]
C. \[\dfrac{{ - 1}}{3}\]
D. \[\dfrac{1}{6}\]

Hint: First, calculate the determinant of the given square matrix \[A\]. Substitute the values in the standard formula of the inverse matrix. In the end, equate the value of the inverse matrix with the given inverse matrix and get the required answer.

Formula used:
Inverse matrix: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]
The adjoint matrix of a \[2 \times 2\] matrix \[A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\] is: \[adj A = \left[ {\begin{array}{*{20}{c}}d&{ - b}\\{ - c}&a\end{array}} \right]\]

Complete step by step solution:
 The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}0&3\\2&0\end{array}} \right]\] and the inverse matrix is \[{A^{ - 1}} = \lambda \left( {adj\left( A \right)} \right)\].

Let’s calculate the determinant of the given matrix \[A\].
\[\left| A \right| = \left( {0 \times 0} \right) - \left( {2 \times 3} \right)\]
\[ \Rightarrow \left| A \right| = - 6\]

We know that, the standard formula for the inverse of a matrix
 \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]
Substitute the given equation on the left-hand side.
\[\lambda \left( {adj\left( A \right)} \right) = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]
Cancel out the common term from both sides.
\[\lambda = \dfrac{1}{{\left| A \right|}}\]
Substitute the value of the determinant in the above equation.
\[\lambda = \dfrac{1}{{ - 6}}\]
\[ \Rightarrow \lambda = \dfrac{{ - 1}}{6}\]
Hence the correct option is A.

Note: Students often get confused and calculate the inverse matrix of the matrix by calculating the co-factors, adjoint matrix, and determinant using the formula of the inverse matrix. This method is correct but too lengthy for this type of question.