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In questions 1.1 to 1.7 below, one or more of the alternatives are correct.  Write the code letter(s) a, b, c, d corresponding to the correct alternative(s) in the answer book.  Marks will be given only if all the correct alternatives have been selected and no incorrect alternative is picked up.

1.1). The eigen vector (s) of the matrix

$$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

(a). $(0,0,\alpha)$

(b). $(\alpha,0,0)$

(c). $(0,0,1)$

(d). $(0,\alpha,0)$
asked in Linear Algebra by Veteran (59.4k points) | 929 views
0
I believe the answer should be only B. It is because if we look at the matrix it shows the transformed 3D space into a line where the z^ vector is transformed into a line replicating x axis. In that scenario the only vector that remains unchanged is (alpha, 0, 0) i.e option B.
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could anyone give the complete procedure ?
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Answer is b only why obtion d is consider here ?
+5
Why d is included in answer?

Here we got eign values as 0,0,0.

Now to get eign vector take  λ=0 in

( A-λI) X =0 or AX=0 -----(i)

here X is eign vector in form of column matrix [X1,X2,X3].

If we multiply as per (i) we get equation as

X1*0+X2*0+ aX3=0

Now for LHS to be 0 , X3 must be 0 . Here X1 and X2 can be assumed any value . As per the given option a and c are wrong.hence b and d satisfy the case
0
what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.

1 Answer

+14 votes
Best answer
Answer: B, D

Eigen values are: 0,0,0

The eigen vector should satisfy the equation: $\alpha$z = 0
answered by Boss (34.1k points)
selected by
0
yes, I also think ans should be option B only.
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how b and d?
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here by normal approach we get in eigen vector x3=0

So, answer will be b) and d)
0
^ Will $-\lambda^3 = 0$ not be the characterastic equations for this matrix?
So how third value came $\alpha$?
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yes (B) and (D) should be correct option!
0

@Manu Thakur

here we have the Basic variable as "z" and free variable as "x" and "y".

So, keeping x = t and y = s, and z = $\alpha$

Then vector corresponding to x is $[1, 0, 0]^T$

And vector corresponding to y is $[0, 1, 0]^T$

Then why it is not $[1, 0, 0]^T$ And $[0, 1, 0]^T$ instead of $[\alpha, 0 , 0]^T$ And $[0, \alpha, 0]^T$?

0
$\alpha$ or 1 doesnot matter

atleast u have to give answer, from options given
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But it should be $[1,0,0]^T$ And $[0,1,0]^T$ even no option matches. ryt??
0
if B & D are true then A should also be true please check once?


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