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Let $M = \begin{bmatrix} a & b &c \\ b &d & e\\ c & e & f \end{bmatrix}$ be a real matrix with eigenvalues 1, 0 and 3. If the eigenvectors corresponding to 1 and 0 are $\left ( 1,1,1 \right )^T$ and  $\left ( 1,-1, 0 \right )^T$ respectively, then the value of 3f is equal to  _______.
in Linear Algebra by Boss (35.2k points) | 267 views
0
Is it 7?
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yes please add the solution
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@Minipanda

plz tell how u solved

I am getting 2 parallel lines , means unsolvable

1 Answer

+7 votes
Best answer

We know that MX= ⋋X where X is the eigen vector correspondig to eigen value ⋋

So for ⋋=1,

$\begin{pmatrix} a &b &c \\ b &d &e \\ c &e &f \end{pmatrix} * \begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix} = 1* \begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix}$

 

a+b+c=1  --> (1)

b+d+e=1  --> (2)

c+e+f=1   --> (3)

For ⋋=0

 

$\begin{pmatrix} a &b &c \\ b &d &e \\ c &e &f \end{pmatrix} * \begin{bmatrix} 1\\ -1 \\ 0 \end{bmatrix} = 0* \begin{bmatrix} 1\\ -1 \\ 0 \end{bmatrix}$

 

a-b=0 => a=b

b-d=0  => b=d

c-e=0 =>c=e

So, a=b=d and c=e

Let us say a=b=d=x and c=e=y

Now the matrix becomes :

$\begin{bmatrix} x & x &y \\ x & x & y\\ y &y & f \end{bmatrix}$

Given eigen values are 1,0,3

Trace of matrix = sum of eigen values

x+x+f=1+0+3 = 4

=>2x+f=4  --> (4)

Also from equ (1)

x+x+y=1 => 2x+y=1 --> (5)

From equ (2)

y+y+f=1 => 2y+f=1 --> (6)

From (4), (5) and (6),

f=1-2y (from (6) )

Putting this in (4)

2x+f=4  => 2x+1-2y=4

=> 2x-2y =3

     2x+y=1 (from (5) )

Solving these two we get y= -2/3

f=1-2y (from (6) )

= 1 - 2(-2/3) = 1 +4/3 = 7/3

3f = 7.

by Boss (22.5k points)
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0

MiNiPanda  can you explain me only this part

Trace of matrix = sum of eigen values

x+x+f=1+0+3 = 4


 

+1

@Magma

The sum of the elements of the main diagonal of a matrix = sum of all eigen values of the matrix

Ref: Theorem 2 here 

 https://www.adelaide.edu.au/mathslearning/play/seminars/evalue-magic-tricks-handout.pdf

 

 

0
thanks :D
0
nice explanation

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