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$\\ A\ is\ n*n \ matrix \ such \ that \ A^2=I \ and B \ is \ n*1 \ real \ \\vector \ then \ Ax=B \ has \\ \\ a) no \ solution \\ b) unique \ solution\\ c) infinitely \ many \ solution\\ d) none$
in Linear Algebra by (176 points)
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https://gateoverflow.in/187777/in-gate-2007-matrix

There is one discussion on this question but not able to understand the explanation.

Please anyone either explain that explanation or please provide a better one.

1 Answer

+1 vote
They’ve said

$A*A=I$

$A=\frac{I}{A}$

$A=I*A^{-1}$

$A=A^{-1}$

As inverse of A exists so $|A|\neq 0$

i.e. A is non-singular matrix

if A is non-singular matrix then there is no dependency among rows(like R1=R2+aR3 etc.). And there are no zero rows in the matrix A

also non-singular matrices are full rank matrix(their rank=n)

Now coming to the Augmented matrix (A|B)

as rank of A=n, so rank of the augmented matrix will be n too irrespective of values in B

when rank of A=rank of (A|B)=number of unknown then there exists unique solution
by (55 points)
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@aditi19

Okay Got it. Thank you

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