3.2k views

In the given matrix $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$ , one of the eigenvalues is 1. The eigenvectors corresponding to the eigenvalue 1 are

1. $\left\{a\left(4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
2. $\left\{a\left(-4,2,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
3. $\left\{a\left(\sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$
4. $\left\{a\left(- \sqrt{2},0,1\right) \mid a \neq 0, a \in \mathbb{R}\right\}$

edited | 3.2k views
+4

Answer: B. $\left \{ a\left ( -4,2,1 \right ) \bigm| a\neq0,a\in\mathbb{R} \right \}$

+15

see this

0

To find an eigenvector corresponding to the eigenvalue 1

Step 1: Plug in the eigenvector to get $A- \lambda I$

$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$

Step 2: Use $(A- \lambda I)x=0$

$\begin{bmatrix} 0 & -1&2 \\ 0&0 &0 \\ 1&2 & 0 \end{bmatrix}$ $\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}=0$

So,

$-y+2z=0$

and

$x+2y=0$ (Only Option B satisfies this)

Now, make as many linearly dependent vectors out of it. Option B has one such.

$\begin{bmatrix} 1-1 & -1 & 2\\ 0& 1-1& 0\\ 1& 2& 1-1 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2\\ 0& 0& 0\\ 1& 2& 0 \end{bmatrix}*\begin{bmatrix} x\\ y\\ z \end{bmatrix}$

$-y + 2z = 0$

$x + 2y = 0$

Now consider each of the triplets as the value of x, y, z and put in these equations the one which satisfies is the answer.

Why so because an eigen vector represents a vector which passes through all the points which can solve these equations.

So, we can observe that only option B is satisfying the equations.

by Active (2.3k points)
edited
0
Why does applying row/column transformation first not giving the answer? Is it not advisable to do so?
0
+1
We know that

$AX=\lambda X$

$AX-\lambda X=0$

$(A-\lambda I) X=0$

In given question $\lambda=1$
+1

Words of caution:

DO NOT use row or column transformations on the matrices, when eigen vectors and values are concerned.

Let z represents the eigenvalues.

And let the given matrix be A (square matrix of order 3 x3)

The characteristic equation for this is :

AX = zX ( X is the required eigenvector )
AX - zX = 0
[ A - z I ] [X] = 0 ( I is an identity matrix of order 3 )

put z = 1 ( because one of the eigenvalue is 1 )

[ A - 1 I ] [X] = 0

The resultant matrix is :

[ 0 -1 2 ] [x1] [0]
| 0 0 0 ] |x2] =|0|
[ 1 2 0 ] |x3] [0]

Multiplying thr above matrices and getting the equations as:

-x2 + 2x3 = 0 ----------------(1)
x1 + 2x2 = 0-----------------(2)

now let x1 = k, then x2 and x3 will be -k/2 and -k/4
respectively.

hence eigenvector X = { (k , -k/2, -k/4) } where k != 0

put k = -4c ( c is also a constant, not equal to zero ),
we get X = { ( -4c, 2c, 1c ) }, i.e. { c ( -4, 2, 1 ) }

Hence option B.
by Boss (10k points)
0

u hav copied geeksforgeeks or geeksforgeeks copied u ???

https://www.geeksforgeeks.org/gate-gate-cs-2015-set-3-question-25/

if A is a Matrix and X is a Eigen Vector then  this condition follow [A] [X] = Eigen Value * [X]

so,

here Eigen value is 1

by Boss (11.7k points)
0
How did u get the X matrix ?? By using options ???
0
yes @pooja
0
ok ...(y)
+4

@puja refer this video for more clear idea

Given matrix, and it's eigen value is 1,

I write $|A-1.I|=$\begin{vmatrix} 0& -1&2 \\ 0&0 &0 \\ 1& 2& 0\end{vmatrix}$$Now I have to find an Eigen-vector(obviously non-zero) such that |A-I|.X=0 and this is \begin{vmatrix} 0 & -1 & 2\\ 0 & 0 &0 \\ 1 & 2 &0 \end{vmatrix}.\begin{bmatrix} a\\ b\\ c \end{bmatrix}=0 Now, you should think what combinations of columns 1,2 and 3 should I take to get the zero vector \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} And, you can agree if I say, -4times the first column +2 times the second column and 1 times the third column will give you the zero vector i.e. -4\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}+ 2\begin{bmatrix} -1\\ 0\\ 2 \end{bmatrix}+$$\begin{bmatrix} 2\\ 0\\ 0 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$so our X becomes X=$\begin{bmatrix} -4\\ 2\\ 1 \end{bmatrix}$And any linear combination of this X would also result in$|A-I|cX=c\{|A-I|.X\}=c.0=0$provided$c \not=0\$ otherwise we would get the zero vector and the zero vector is a trivial solution to AX=0.

by Boss (30.6k points)