Question

If $A=\left( \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right)$ and $B=\left( \begin{matrix}
   0 & -i \\
   i & 0 \\
\end{matrix} \right)$ then ${{(A+B)}^{2}}$ is equal to?
A . ${{A}^{2}}+{{B}^{2}}$
B. ${{A}^{2}}+{{B}^{2}}+2AB$
C. ${{A}^{2}}+{{B}^{2}}+AB-BA$
D. None of these

Answer 1

Hint: Given two matrices A and B of order $(2\times 2)$. We have to find AB. In order to multiply the matrices, both the matrix are of the same order. As the given matrices are of the same order, so we multiply them and then add the matrices to get the desirable answer.

Complete step by step Solution:
Given matrix is $A=\left( \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right)$ and $B=\left( \begin{matrix}
   0 & -i \\
   i & 0 \\
\end{matrix} \right)$
In matrix A, there are 2 rows and 2 columns.
Similarly in matrix B, there are 2 rows and 2 columns.
So both are of order $(2\times 2)$ matrix.
The basic representation of $(2\times 2)$ matrix is $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
Then AB will represent as $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
AB = $\left( \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right)$$\left( \begin{matrix}
   0 & -i \\
   i & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
The first digit of the first row is the sum of the multiplication of the first row of matrix A and the first column of matrix B.
That is ${{a}_{11}}=0\times 0+1\times i$
Then ${{a}_{11}}=i$
Similarly, for the second digit of the first row, it is the sum of the multiplication of the first row of matrix A with the second column of matrix B.
That is ${{a}_{12}}=0\times -i+1\times 0$
Then ${{a}_{12}}=0$
Now for the second row
The first digit of the second row is the sum of the multiplication of the second row of matrix A with the first column of matrix B
That is ${{a}_{21}}=1\times 0+0\times i$
Then ${{a}_{21}}=0$
Similarly, for the second digit of the first row, it is the sum of the multiplication of the second row of matrix A with the second column of matrix B.
That is ${{a}_{22}}=1\times -i+0\times 0$
Then ${{a}_{22}}=-i$
Now we substitute all the values in the AB matrix and get
AB = $\left( \begin{matrix}
   i & 0 \\
   0 & -i \\
\end{matrix} \right)$
Thus the product of A and B is $\left( \begin{matrix}
   i & 0 \\
   0 & -i \\
\end{matrix} \right)$
Similarly, we find out BA
BA =$\left( \begin{matrix}
   0 & -i \\
   i & 0 \\
\end{matrix} \right)$$\left( \begin{matrix}
   0 & 1 \\
   1 & 0 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
${{a}_{11}}=0\times 0+-i\times 1=-i$
${{a}_{12}}=0\times 1+-i\times 0=0$
${{a}_{21}}=i\times 0+0\times 1=0$
${{a}_{22}}=i\times 1+0\times 0=i$
Now we substitute all the values in the BA matrix and get
BA = $\left( \begin{matrix}
   -i & 0 \\
   0 & i \\
\end{matrix} \right)$ = - AB
Since AB + BA= O
Therefore,
${{(A+B)}^{2}}={{A}^{2}}+{{B}^{2}}$

Therefore, the correct option is (A).

Note: Remember that while multiplying the matrices the elements (that is ) ${{a}_{11}},{{a}_{12}}$ represents the row number and column number. For example, ${{a}_{11}}$ tells us about the multiplication of the first row of matrix 1 with the first column of matrix 2. ${{a}_{12}}$ represents the multiplication of the first row of matrix 1 with the second column of matrix 2 and so on.