Question

If two non – singular matrices $A$ and $B$ of same order obeys \[A\,B\, = \,B\,A = I\], then $B$ is
$
  A\,)\,A\, = \,{B^{ - 1}} \\
  B\,)\,B\, = \,A \\
  \,C\,)\,B\, = \,{A^{ - 1}} \\
  \,D\,)\,A\, = \,B \\
 $

Answer 1

Hint:
Split the given equation in two parts, and equate both the equations separately. Multiply ${A^{ - 1}}$ and ${B^{ - 1}}$ correspondingly to the splitted equation, to find the value of $A$ and $B$.

Useful formula: The basic formula which is used in this problem is $A\,{A^{ - 1\,}}\, = \,1$ as well as $B\,{B^{ - 1}}\, = \,1$. Another formula which is used in the problem is ${A^{ - 1}}\,I\, = \,{A^{ - 1}}$ as like that $B{\,^{ - 1}}\,I\, = \,{B^1}$.

Complete step by step solution:
Given that, $A$ and $B$ are two non – singular matrices. Both $A$ and $B$ obeys the order\[A\,B\, = \,B\,A = I\].
Now, we want to find the order $B$ if it obeys the given order.
If $A$ and $B$ are two non – singular matrices, then
\[A\,B\, = \,B\,A = I\]
Now, split the above equation in two parts as follows,
\[A\,B\, = \,I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,(\,1\,)\]
 $B\,A\, = \,I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,(\,2\,)$
Now, consider equation$(\,2\,)$:
$B\,A\, = \,I$
The above carries $B$ as the initial value, so multiply the above equation with ${B^{ - 1}}$ to get the value of $A$:
Multiply the above equation with ${B^{ - 1}}$ on both sides of the equation:
${B^{ - 1}}\,B\,A\, = \,{B^{ - 1}}\,I$
By using the formula, ${B^{ - 1}}\,I\, = \,{B^{ - 1}}$ substitute the relation in above equation:
${B^{ - 1}}\,B\,A = \,{B^{ - 1}}\,$
By using the formula $B\,{B^{ - 1}}\, = \,1$, the above will become as follows:
$(\,1\,)\,A\, = \,{B^{ - 1}}$
Thus, the value of $A$ is
 $A\, = \,{B^{ - 1}}$
As like equation $(\,2\,)$, solve the equation $(\,1\,)$ as follows:
The above carries $A$ as the initial value, so multiply the above equation with ${A^{ - 1}}$to get the value of $B$:
Multiply the above equation with ${A^{ - 1}}$ on both sides of the equation:
 \[{A^{ - 1}}\,A\,B\,\, = \,{A^{ - 1}}\,I\]
By using the formula, ${A^{ - 1}}\,I\, = \,{A^{ - 1}}$ substitute the relation in above equation:
${A^{ - 1}}A\,B\, = \,{A^{ - 1}}$
By using the formula, the above will become as follows:
$(\,1\,)\,B\, = \,{A^{ - 1}}$
Thus, the value of $B$ is
 $B\, = \,{A^{ - 1}}$
By solving the given equation of \[A\,B\, = \,B\,A = I\], we have found the value of $A$ and $B$:
$A\, = \,{B^{ - 1}}$
$B\, = \,{A^{ - 1}}$
With the help of above given equation, we have found the value of $B$ as
$B\, = \,{A^{ - 1}}$

Thus, the option is the answer.

Note:
Multiply of the variable and its inverse should always be $1$ and the multiply of inverse of variable and $I$ should always be inverse of variable. To find the value of $B$ multiply the equation $(\,2\,)$with ${A^{ - 1}}$.