The matrix \[\left[ \begin{matrix}
   \lambda & 7 & -2 \\
   4 & 1 & 3 \\
   2 & -1 & 2 \\
\end{matrix} \right]\] is a singular matrix if \[\lambda \] is
\[\begin{align}
  & \left( \text{A} \right)\dfrac{2}{5} \\
 & \left( \text{B} \right)\dfrac{5}{2} \\
 & \left( \text{C} \right)\text{-5} \\
 & \left( \text{D} \right)\text{none of these} \\
\end{align}\]

Hint: These types of problems are pretty straight forward and are very easy to solve. For the given type of problems we first need to understand what a singular matrix means. We also need to remember how to find the determinant of a given matrix. A singular matrix is a matrix whose value of the determinant is \[0\] . We already know that the determinant of a matrix is defined if and only if the given matrix is a square matrix.

Complete step-by-step answer:
Now we start off with our solution as,
From the given matrix in our problem we can very easily find the order of the matrix as \[3\times 3\] , as the matrix contains \[3\] rows and \[3\] columns. Now we need to find out the determinant of the given square matrix and then equate it to zero. The determinant is calculated as,
\[\left| \begin{matrix}
   \lambda & 7 & -2 \\
   4 & 1 & 3 \\
   2 & -1 & 2 \\
\end{matrix} \right|\]
The first term is positive, the second term is negative and the third term is positive. Finding by using the first row, we write,
\[\lambda \left( 1\cdot 2-\left( -1\cdot 3 \right) \right)-7\left( 4\cdot 2-3\cdot 2 \right)+\left( -2 \right)\left( 4\cdot \left( -1 \right)-1\cdot 2 \right)\]
Now, taking out the negative signs we get,
\[\Rightarrow \lambda \left( \left( 1\cdot 2 \right)+\left( 1\cdot 3 \right) \right)-7\left( \left( 4\cdot 2 \right)-\left( 3\cdot 2 \right) \right)-2\left( \left( 4\cdot -1 \right)-\left( 1\cdot 2 \right) \right)\]
Now doing the necessary multiplications we get,
\[\Rightarrow \lambda \left( 2+3 \right)-7\left( 8-6 \right)-2\left( -4-2 \right)\]
Adding and Subtracting, we get,
\[\Rightarrow \lambda \left( 5 \right)-7\left( 2 \right)-2\left( -6 \right)\]
Now, evaluating it,
\[\begin{align}
  & \Rightarrow 5\lambda -14+12 \\
 & \Rightarrow 5\lambda -2 \\
\end{align}\]
Now, equating the above intermediate equation to zero (as it is given to be a singular matrix) we get,
\[5\lambda -2=0\]
Evaluating for the value of \[\lambda \] we get,
\[\begin{align}
  & 5\lambda =2 \\
 & \Rightarrow \lambda =\dfrac{2}{5} \\
\end{align}\]
Now, from the options given to our problem, we see that our found out answer matches to the first option i.e. option (A).
So, the correct answer is “Option A”.

Note: We must always remember that a matrix should always have the same number of rows and columns, or in other words the matrix should be a square matrix, to be able to find the determinant of the matrix. We should also never forget the steps to calculate the determinant of a square matrix. For singular matrices, we should keep in mind that the determinant always evaluates to zero.