Question

If A =\[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\], B = \[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\] where \[i = \sqrt { - 1} \], then the correct relation is
A. \[A + B = 0\]
B. \[{A^2} = {B^2}\]
C. \[A - B = 0\]
D. \[{A^2} + {B^2} = 0\]

Answer 1

Hint: In this question, we have given two matrices A and B and we have to find out the correct option which satisfies the given matrix. First we add the given matrix then we check whether the answer is zero matrix or not. Then we square both the matrix and check it whether it is equal or not. Then in third option, we subtract the matrix and in option fourth, we add the square of both the matrix and check the correct option.

Formula Used :
Here in this question we have to use the concept of multiplication of matrices.

Complete Step by Step Solution:
We have given the matrix A = \[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\] and B = \[\]\[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\]
We solve it by using options
Let us consider Option (1)
We check whether \[A + B = 0\]
\[A + B\]= \[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\] + \[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\]
\[A + B\]= \[\left[ {\begin{array}{*{20}{c}}{i + 0}&{0 + i}\\{0 + i}&{ - i + 0}\end{array}} \right]\]
Solving further, we get
\[A + B\]= \[\left[ {\begin{array}{*{20}{c}}i&i\\i&{ - i}\end{array}} \right]\]
Hence \[A + B\]is a non- zero matrix.
Thus \[A + B \ne 0\]
Now we consider Option (2)
In Option (2), first we find \[{A^2}\]and \[{B^2}\]
\[{A^2}\]= \[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\]
Multiplication of both the matrices, we get
\[{A^2}\]= \[\left[ {\begin{array}{*{20}{c}}{i \times i + 0 \times 0}&{i \times 0 + 0 \times - i}\\{0 \times i - i \times 0}&{0 \times 0 - i \times i}\end{array}} \right]\]
\[{A^2}\]= \[\left[ {\begin{array}{*{20}{c}}{{i^2}}&0\\0&{{i^2}}\end{array}} \right]\]
As \[i = \sqrt { - 1} \]
So \[{i^2} = - 1\]
Hence \[{A^2}\]= \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\]
Similarly \[{B^2}\]= \[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\] \[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\]
On multiplying both the matrix, we get
\[{B^2}\]= \[\left[ {\begin{array}{*{20}{c}}{0 \times 0 + i \times i}&{i \times 0 + 0 \times i}\\{0 \times i + i \times 0}&{0 \times - 0 + i \times i}\end{array}} \right]\]
Then \[{B^2}\]= \[\left[ {\begin{array}{*{20}{c}}{{i^2}}&0\\0&{{i^2}}\end{array}} \right]\]
Hence \[{B^2}\]= \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\]
Hence \[{A^2} = {B^2}\]
Now we check Option ( C )
\[A - B\]= \[\left[ {\begin{array}{*{20}{c}}i&0\\0&{ - i}\end{array}} \right]\]- \[\left[ {\begin{array}{*{20}{c}}0&i\\i&0\end{array}} \right]\]
\[A - B\]= \[\left[ {\begin{array}{*{20}{c}}{i - 0}&{0 - i}\\{0 - i}&{ - i - 0}\end{array}} \right]\]
Hence \[A - B\]= \[\left[ {\begin{array}{*{20}{c}}i&{ - i}\\{ - i}&{ - i}\end{array}} \right]\]
Thus \[A - B \ne 0\]
Now we check Option (D)
As we find the values of \[{A^2}\]and \[{B^2}\] in the Option (B)
We put the values directly
Then \[{A^2} + {B^2}\]= \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\] + \[\left[ {\begin{array}{*{20}{c}}{ - 1}&0\\0&{ - 1}\end{array}} \right]\]
\[{A^2} + {B^2}\]= \[\left[ {\begin{array}{*{20}{c}}{ - 2}&0\\0&{ - 2}\end{array}} \right]\]
Hence, we see that only Option ( B ) satisfies the given equation.

Thus Option (B) is correct :

Note: In these types of questions, students made mistakes while the multiplication of two matrices. In multiplying the matrices, the number of columns of the first matrix must be equal to the number of rows of the second matrix. When we multiply the matrices, then the parts of the rows in the first matrix are multiplied with the columns in the second matrix.