For the matrix \[A = \left[ {\begin{array}{*{20}{c}}
  1&1&0 \\
  1&2&1 \\
  2&1&0
\end{array}} \right]\]which of the following are correct
A. \[{A^3} + 3{A^2} - I = 0\]
B. \[{A^3} - 3{A^2} - I = 0\]
C. \[{A^3} + 2{A^2} - I = 0\]
D. \[{A^3} - {A^2} + I = 0\]

Hint: In the given problem we first find the value of \[{A^2}\]. Similarly, we find the value of \[{A^3}\] by multiplying \[{A^2}\] with \[A\]. We then perform different operations with \[{A^2}\] and \[{A^3}\] in accordance with the given options to get the correct option.

Formula used:
If \[A = {[{a_{ij}}]_{m \times n}}\] and \[B = {[{b_{ij}}]_{n \times p}}\] then we can say that \[A \times B = C\] where the value of C is
\[C = {[{c_{ij}}]_{m \times p}}\]
Here \[{c_{ij}} = \mathop \sum \limits_{j = 1}^m {a_{ij}}{b_{jk}} = {a_{i1}}{b_{1k}} + {a_{i2}}{b_{2k}} + ........ + {a_{im}}{b_{mk}}\]

Complete step by step solution:
We are given that,
\[A = \left[ {\begin{array}{*{20}{c}}
  1&1&0 \\
  1&2&1 \\
  2&1&0
\end{array}} \right]\]
On squaring A, we get \[{A^2}\]
\[{A^2} = A.A = \left[ {\begin{array}{*{20}{c}}
  1&1&0 \\
  1&2&1 \\
  2&1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&1&0 \\
  1&2&1 \\
  2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  2&3&1 \\
  5&6&2 \\
  3&4&1
\end{array}} \right]\]
On multiplying the scalar 3 with \[{A^2}\]we get,
\[3{A^2} = 3\left[ {\begin{array}{*{20}{c}}
  2&3&1 \\
  5&6&2 \\
  3&4&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  6&9&3 \\
  {15}&{18}&6 \\
  9&{12}&3
\end{array}} \right]\]
On multiplying \[{A^2}\] with A we get \[{A^3}\]
\[{A^3} = {A^2}.A = \left[ {\begin{array}{*{20}{c}}
  2&3&1 \\
  5&6&2 \\
  3&4&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&1&0 \\
  1&2&1 \\
  2&1&0
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  7&9&3 \\
  {15}&{19}&6 \\
  9&{12}&4
\end{array}} \right]\]
Therefore \[{A^3} - 3{A^2} = \left[ {\begin{array}{*{20}{c}}
  7&9&3 \\
  {15}&{19}&6 \\
  9&{12}&4
\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}
  6&9&3 \\
  {15}&{18}&6 \\
  9&{12}&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = I\]
\[
    \\
   \Rightarrow {A^3} - 3{A^2} - I = 0 \\
 \]
Option B. is the correct answer.

Note: To solve the given problem, one must know to add, subtract and multiply two matrices. One must make sure that the terms are added before giving the resultant value in each position of the resultant matrix during matrix multiplication.