If matrix A=\[\left[{\begin{array}{*{20}{c}}1&{ - 1}\\1&1\end{array}} \right]\], then
A. \[{A^T} =
\left[ {\begin{array}{*{20}{c}}1&1\\1&{ - 1}\end{array}} \right]\]
B. \[{A^{ - 1}} =
\left[ {\begin{array}{*{20}{c}}1&1\\{ - 1}&1\end{array}} \right]\]
C. \[A\left[
{\begin{array}{*{20}{c}}1&1\\{ - 1}&1\end{array}} \right] = 2I\]
D. \[\lambda A =
\left[ {\begin{array}{*{20}{c}}\lambda &{ - \lambda }\\1&{ -
1}\end{array}} \right]\] where \[\lambda\] is a non-zero scalar.

Hint:

First of all find the transpose of matrix A and check each option one by one

Formula Used:
For the multiplication of two 2×2 matrices, we will solve like this
Let a matrix of 2×2 be \[\left[
{\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}}
\right]\]and another 2×2 matrix be \[\left[
{\begin{array}{*{20}{c}}{{b_{11}}}&{{b_{12}}}\\{{b_{21}}}&{{b_{22}}}\end{array}}
\right]\]
So by multiplication of these matrices, we get
\[\left[
{\begin{array}{*{20}{c}}{{a_{11}}{b_{11}} +
{a_{12}}{b_{21}}}&{{a_{11}}{b_{12}} + {a_{12}}{b_{22}}}\\{{a_{21}}{b_{11}}
+ {a_{22}}{b_{21}}}&{{a_{21}}{b_{12}} + {a_{22}}{b_{22}}}\end{array}}
\right]\]
And if we want to transpose a given matrix we simply exchange rows
to columns and columns to rows. We will get the transpose of a given matrix.
And if we want to find\[{A^{ -
1}}\], it has the formula \[{A^{
- 1}} = \frac{{adjA}}{{\left| A \right|}}\]




Complete Step-By-Step Solution:

First of all, we will see that if the given matrix A= \[\left[ {\begin{array}{*{20}{c}}1&{ -
1}\\1&1\end{array}} \right]\] is equal to its transpose or not.
\[{A^T} = \left[ {\begin{array}{*{20}{c}}1&1\\{ -1}&1\end{array}} \right]\].
But this is not equal to the given matrix \[\left[{\begin{array}{*{20}{c}}1&{ - 1}\\1&1\end{array}} \right]\].
So, option A is not correct.
Now we will see that whether \[A\left[ {\begin{array}{*{20}{c}}1&1\\{ -
1}&1\end{array}} \right] = 2I\]
So for that, we have to multiply the given matrix A=\[\left[ {\begin{array}{*{20}{c}}1&{ -
1}\\1&1\end{array}} \right]\] by \[\left[ {\begin{array}{*{20}{c}}1&1\\{ -1}&1\end{array}} \right]\]
So, product of these two matrices is \[\left[ {\begin{array}{*{20}{c}}{1(1) + ( - 1)( -
1)}&{1(1) + ( - 1)(1)}\\{1(1) + (1)( - 1)}&{1(1) + (1)(1)}\end{array}}\right]\]
\[\left[{\begin{array}{*{20}{c}}{1 + 1}&{1 - 1}\\{1 - 1}&{1 + 1}\end{array}}\right]\]
\[\left[{\begin{array}{*{20}{c}}2&0\\0&2\end{array}} \right]\]
Now we can clearly see that it is equal to \[2\left[{\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\] that is \[2I\]
where \[I\] is an identity matrix.
\[A\left[ {\begin{array}{*{20}{c}}1&1\\{- 1}&1\end{array}} \right] = 2I\]
Hence, Option C is correct.

Note:
In this question, we have to remember that if such a question comes we have to solve all the parts and see which part is satisfying our condition. Like in this one only we have to solve and then and
then the product of the matrix which is satisfying the condition.