Question

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

Answer 1

Hint: Here \[I\] is an identity matrix or unit matrix (All the non-diagonal elements are zero and elements of diagonal are 1). We know that a non-singular matrix means that whose determinant is not zero. Also the order of the two matrices are the same. By multiplying any matrix by identity matrix, gives the matrix itself. Multiplication of matrix and its inverse matrices we will get an identity matrix. Using these properties we can solve the above problem.

Complete step by step solution:
Given,
\[AB = BA = I\].
Thus we have,
\[\Rightarrow BA = I{\text{ }} - - - (1)\]
\[\Rightarrow AB = I{\text{ }} - - - (2)\].
Now take equation (1)
\[\Rightarrow BA = I\]
Multiply by \[{B^{ - 1}}\] on both sides we have,
\[\Rightarrow {B^{ - 1}}.(BA) = {B^{ - 1}}.I\]
\[\Rightarrow \left( {{B^{ - 1}}B} \right).A = {B^{ - 1}}I\]
By multiplying any matrix by identity matrix, gives the matrix itself, that is \[ \Rightarrow A.I = A\].
Multiplication of matrix and its inverse matrices we will get an identity matrix, that is \[ \Rightarrow A.{A^{ - 1}} = I\]. Using this property in above we have,
\[\Rightarrow I.A = {B^{ - 1}}I\]
\[ \Rightarrow A = {B^{ - 1}}\].
Now take equation (2),
\[\Rightarrow AB = I\]
Multiply by \[{A^{ - 1}}\] on both sides we have,
 \[\Rightarrow {A^{ - 1}}\left( {AB} \right) = {A^{ - 1}}.I\]
\[\Rightarrow \left( {{A^{ - 1}}A} \right)B = {A^{ - 1}}.I\]
Applying the identity properties as done above we have,
\[\Rightarrow I.B = {A^{ - 1}}.I\]
\[ \Rightarrow B = {A^{ - 1}}\].
Thus the required answer is \[A = {B^{ - 1}}\] and \[B = {A^{ - 1}}\].

Hence the correct answer is option ‘C’.

Note: Keep in mind that the above solution satisfies only if the order of the two matrices are same. That is, if a matrix has ‘m’ rows and ‘n’ columns then the order of the matrix is given by \[m \times n\]. If matrix A is of order \[m \times n\] then order of B is also \[m \times n\].