If \[\begin{array}{*{20}{c}}A& = &{\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&2\end{array}} \right]}\end{array}\] and I is the unit matrix of order 2, then \[{A^2}\]
A. \[4A - 3I\]
B. \[3A - AI\]
C. \[A - I\]
D. \[A + I\]

Hint: In this question, first of all, we will determine the square of the given matrix. And then we will split the square matrix into two different matrices such that one is a matrix \[A\] and another is the unit matrix. We know that the unit matrix is that matrix in which the elements of the principal diagonal should be one and other elements should be zero. After that, we will get a linear equation; hence, we will get a suitable answer.

Formula used:
⟹\[{A^2} = A \times A\]

Complete step by step Solution:
As we have given a matrix such as,
\[\begin{array}{*{20}{c}}{ \Rightarrow A}& = &{\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&2\end{array}} \right]}\end{array}\]
Now we will determine the square of the given matrix \[A\]. Therefore, we will write
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{A \times A}\end{array}\]
Now we will put the matrix in the above expression. Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&2\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&2\end{array}} \right]}\end{array}\]
And then we will get,
\[{A^2} = \left( {\begin{array}{*{20}{c}}{4 + 1}&{ - 2 - 2}\\{ - 2 - 2}&{1 + 4}\end{array}} \right)\]
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{\left[ {\begin{array}{*{20}{c}}5&{ - 4}\\{ - 4}&5\end{array}} \right]}\end{array}\]
Now we will split the above matrix in such a manner that one matrix will be the base matrix (Matrix \[A\]) and another matrix will be a unit matrix. Therefore,
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{\left[ {\begin{array}{*{20}{c}}8&{ - 4}\\{ - 4}&8\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right]}\end{array}\]
Now
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{4\left[ {\begin{array}{*{20}{c}}2&{ - 1}\\{ - 1}&2\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]}\end{array}\]
Now we know that in the above expression, we have the matrix \[A\] and unit matrix \[I\]. Therefore, we will write,
\[ \Rightarrow \begin{array}{*{20}{c}}{{A^2}}& = &{4A - 3I}\end{array}\]

Therefore, the correct option is (A).

Note:In this question, the first point is to keep in mind that the unit matrix has the elements of the principal diagonal are one, and the other elements of the matrix zero. Students should keep in mind that matrices have zero divisors and can be easily solved but in some cases when there is matrix multiplication it should be solved cautiously and it should be kept in mind that multiply each element of the first matrix's column by each element of the second matrix's rows and add them all.