+16 votes
1.8k views

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)

1. $(0,0,\alpha)$
2. $(\alpha,0,0)$
3. $(0,0,1)$
4. $(0,\alpha,0)$
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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 ?
+15
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
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what about (alpha,alpha,0) vectors? will they be eigen vectors too.I think they are.
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Bind blowing explanation satyajeet sir
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Yes they will be eigen [email protected] gow

## 3 Answers

+12 votes
Best answer
Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$

So the question asks

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.
answered by Boss (25.4k points)
selected
0
The second part of the answer is difficult for me to grasp.
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Which one?
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From the statement "look carefully"
+5
ohh that one. See from this example.

Suppose you have a system like

$\begin{bmatrix} 1 &3 &6 \\ 7 & 5& 9\\ 3 &8 & 2 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$

this is equivalent to

$x_1\begin{bmatrix} 1\\ 7\\ 3 \end{bmatrix}+x_2\begin{bmatrix} 3\\ 5\\ 8 \end{bmatrix}+x_3\begin{bmatrix} 6\\ 9\\ 2 \end{bmatrix}$

In the question given, you have to produce a zero vector as result. What suitable combinations of $x_1,x_2,x_3$ can you take?
+15 votes
Answer: $B, D$

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$
answered by Boss (33.8k points)
edited
0
yes, I also think ans should be option B only.
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how b and d?
0
here by normal approach we get in eigen vector x3=0

So, answer will be b) and d)
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^ 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

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
0
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?
0
Ok
+4 votes
Given matrix is an upper triangular matrix. Therefore diagonal elements of A are eigen values of A, i.e. λ=0,0,0.

Consider $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Put λ=0

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have

$\alpha$$x_3=0 \Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.

Hence in the given options $B$ and $D$ are correct.
answered by Active (4.8k points)

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