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Let A be an nxn real matrix such that A^2=I and y be an n-dimensional vector. Then the linear system of equations AX=Y has

A) No solution

B) Unique Solution

C) More than one but finitely many independent solutions

D) infinitely many independent solutions

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Unquie solution i.e exactly one solution which is $Y \times A$
Given condition $A^2=I \Rightarrow A \times A=I \Rightarrow A=A^{-1}$

Now Y is $n$ dimensional vector .

$\therefore AX=Y \\ \Rightarrow X_{n \times 1}= A^{-1}_{n \times n} \times Y_{n \times 1} \\ \Rightarrow X=AY(solution)$
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Unique solution is the right answer.

@saxena0612
To check whether there exists unique solution, the rank of (A|Y) = rank of (A). Since A inverse exists hence |A| is not equal to zero. So rank =n.
Next, rank= no. Of unknowns=n. Hence unique. Is this the correct approach to deduce?
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I didn`t get your solution but Determinant of $A \ is \neq0$ how it implies that when you will create augument matrix $AY$ it will result the same rank which will be equal to no of unknowns?
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If |A| ≠0 then rank of A = n. Since X is a vector of n dimension so I took the no. Of unknowns as n. As it equal to the rank so unique solution exists.

If I am wrong then please correct it and also say how to solve it.
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@MiNiPanda, U r right. since |A| ≠0 so rank of (A|Y) = rank of (A) , Y is a n-dimensional vector. hence unique solution exist only.