Question

If A and B are two square matrices such that $ B = - {A^{ - 1}}BA $ , then $ {\left( {A + B} \right)^2} $ is equal to ?

Answer 1

Hint: In order to determine the value of $ {\left( {A + B} \right)^2} $ ,multiply the both side of the given equation $ B = - {A^{ - 1}}BA $ with the $ A $ . Remember that the multiplication of a matrix with its inverse gives the identity matrix. Now expand the $ {\left( {A + B} \right)^2} $ as $ \left( {A + B} \right)\left( {A + B} \right) $ .simplify it further and use the result obtained earlier in this expansion to get the required result.

Complete step-by-step answer:
We are given two square matrices A and B having relation between as $ B = - {A^{ - 1}}BA $
Since, $ B = - {A^{ - 1}}BA $
Let's multiply $ A $ on both sides of the equation, we get
 $ AB = - A{A^{ - 1}}BA $
As we know when a matrix is multiplied with its inverse matrix, it results into identity matrix. So here $ A{A^{ - 1}} = I $
 $ AB = - IBA $
 $ AB = - BA $ ------(1)
We have to find the value of $ {\left( {A + B} \right)^2} $ , lets rewrite this thing
 $
\Rightarrow {\left( {A + B} \right)^2} = \left( {A + B} \right)\left( {A + B} \right) \\
\Rightarrow {\left( {A + B} \right)^2} = A.A + A.B + B.A + B.B \;
  $
From the equation (1) $ AB = - BA $ ,putting this into the above expression we get
 $
\Rightarrow {\left( {A + B} \right)^2} = {A^2} - BA + BA + {B^2} \\
\Rightarrow {\left( {A + B} \right)^2} = {A^2} + {B^2} \;
  $
Therefore the $ {\left( {A + B} \right)^2} $ is equal to $ {A^2} + {B^2} $ .
So, the correct answer is “ $ {A^2} + {B^2} $ ”.

Note: Square Matrix: Square matrix is the matrix which has no rows equal to the number of columns in the matrix. The order of the square matrix is $ n \times n $ where n is any natural number.
1.The number of rows and columns are same for the original matrix \[\;\left[ B \right]\] as that of inverse of that same matrix \[\;{\left[ B \right]^{ - 1}}\;\].