GO Classes CS/DA 2025 | Weekly Quiz 4 | Linear Algebra | Question: 6

Answer 1

 $Given  Ax = \lambda x$

As A in invertible so

Option A:  $A^{-1}Ax = \lambda A^{-1}x $ (multiply by A^{-1) on both side)

 (1/ $\lambda$)x = $A^{-1}$x

$A^{-1}$x = $\lambda^{‘}$x 
Hence option A is true 

Option B:  $AAx =  \lambda A x $ (multiply by A on both side)

$A^{2}x = \lambda^{2}x$ (as Ax =$\lambda x$)

Finally $A^{2}x = \lambda^{‘}x$
hence B is true.
Option C:   $Ax + x = \lambda x + x $ (Adding x on both side)

$(A +I)x = (\lambda +1)x$

$(A +I)x = \lambda^{‘’}x$

so C is Also True.
Option D:similar to option C just add 2x on both side.
 

$Hence A,B,C,D all are true$

 

 

Comments

if A is not invertable what all are correct options?

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In think then only option A will be false as inverse of A is not possible. Other options are not depends on A inverse.

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@RK301112 Then, atleast one eigenvalue will be 0, so lambda can't not go in denominator as done in above derivation. 

Except option A,

All other option are correct.