If the matrix $\left[ {\begin{array}{*{20}{c}}
  0&1&{ - 2} \\
  { - 1}&0&3 \\
  \lambda &{ - 3}&0
\end{array}} \right]$is singular, then $\lambda = $
A. $ - 2$
B. $ - 1$
C. $1$
D. $2$

Hint: In this question, we have to check whether the given matrix is singular or not. To determine that, we must know the properties of a singular matrix, that is, a matrix is said to be singular when its determinant is equal to zero and how to find the determinant of the matrix. To find the answer to the given problem, we will find the determinant of the given matrix.

Formula Used:
1. In order to find the determinant of \[3 \times 3\]matrix, the formula used is given below:
\[A = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right]\]
\[\left| A \right| = \]\[{a_{11}}{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right| + {a_{12}}{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{33}}}
\end{array}} \right| + {a_{13}}{( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{22}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right|\]

2. In order to find the determinant of \[2 \times 2\]matrix, the formula used is given below:
\[M = \left[ {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}} \\
  {{a_{21}}}&{{a_{22}}}
\end{array}} \right]\]
\[\left| M \right| = {a_{11}}{a_{22}} - {a_{12}}{a_{21}}\]

Complete step by step Solution:
Assume the given matrix as \[A = \left[ {\begin{array}{*{20}{c}}
  0&1&{ - 2} \\
  { - 1}&0&3 \\
  \lambda &{ - 3}&0
\end{array}} \right]\]
Now, identify all the elements of the matrix\[A\]as per the given determinant formula.
Here,
\[{a_{11}} = 0,\]\[{a_{12}} = 1,\]\[{a_{13}} = - 2,\]\[{a_{21}} = - 1,\]\[{a_{22}} = 0,\]\[{a_{23}} = 3,\]\[{a_{31}} = \lambda ,\]\[{a_{32}} = - 3,\]\[{a_{33}} = 0\]
Substitute all the values in the determinant formula, as shown below:
\[\left| A \right| = \]\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  0&3 \\
  { - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  { - 1}&3 \\
  \lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  \lambda &{ - 3}
\end{array}} \right|\]
In order that the given matrix\[A\]to be singular, the determinant of\[A\]should be equal to zero.
\[\left| A \right| = 0\]
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  0&3 \\
  { - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  { - 1}&3 \\
  \lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  \lambda &{ - 3}
\end{array}} \right| = 0\]
Consider the above equation as \[eq(1)\].
For solving the above equation, use the 2nd formula, mentioned at the start of the solution, to find the determinant of \[2 \times 2\]matrix.
\[\left| {\begin{array}{*{20}{c}}
  0&3 \\
  { - 3}&0
\end{array}} \right| \Rightarrow (0 \times 0) - (3 \times ( - 3)) = 0 - ( - 9) = 9\]
\[\left| {\begin{array}{*{20}{c}}
  { - 1}&3 \\
  \lambda &0
\end{array}} \right| \Rightarrow (( - 1) \times 0) - (3 \times \lambda ) = 0 - 3\lambda = - 3\lambda \]
\[\left| {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  \lambda &{ - 3}
\end{array}} \right| \Rightarrow (( - 1) \times ( - 3)) - (0 \times \lambda ) = 3 - 0 = 3\]
On substituting the above values in the\[eq(1)\], we get:
\[0{( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  0&3 \\
  { - 3}&0
\end{array}} \right| + 1{( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  { - 1}&3 \\
  \lambda &0
\end{array}} \right| + ( - 2){( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  \lambda &{ - 3}
\end{array}} \right| = 0\]
\[[0{( - 1)^{1 + 1}} \times 9] + [1{( - 1)^{1 + 2}} \times ( - 3\lambda )] + [( - 2){( - 1)^{1 + 3}} \times 3] = 0\]
Adding the powers,
\[[0{( - 1)^2} \times 9] + [1{( - 1)^3} \times ( - 3\lambda )] + [( - 2){( - 1)^4} \times 3] = 0\]
\[[0(1) \times 9] + [1( - 1) \times ( - 3\lambda )] + [( - 2)(1) \times 3] = 0\]
On solving further, we get:
\[[0 \times 9] + [( - 1) \times ( - 3\lambda )] + [( - 2) \times 3] = 0\]
\[0 + 3\lambda + ( - 6) = 0\]
On simplifying the equation, we get:
\[3\lambda - 6 = 0\]
Taking -6 to the RHS,
\[3\lambda = 6\]
Taking 3 to the RHS and dividing by 6,
\[\lambda = 2\]
Now, it is clear that, in order for the matrix \[A\] to be singular, \[\lambda \] should be equal to 2.

Therefore, the correct option is (D).

Additional Information: In this type of question, where we have to check whether the given matrix is singular or not, we can use another method using the concept of minors and co-factors.
The minor of an element \[({a_{ij}})\] of a square matrix of any order is the determinant of the matrix that is obtained by removing the row (\[{i^{th}}\]row) and the column (\[{j^{th}}\]column) containing the element.
The co-factor of an element \[({a_{ij}})\] of a square matrix of any order is its minor multiplied by \[{( - 1)^{^{i + j}}}\].
Co-factor an element = \[{( - 1)^{i + j}} \times \](minor of the element)

Note: Since, in this question, we have to check whether the given matrix is singular or not, always keep in mind to find the determinant of the given matrix. If the determinant of the matrix is not equal to zero, then the matrix is a non-singular matrix.