GATE CSE 2017 Set 2 | Question: 52

Question

If the characteristic polynomial of a  $3 \times 3$  matrix $M$ over $\mathbb{R}$ (the set of real numbers) is $\lambda^3 – 4 \lambda^2 + a \lambda +30, \quad a \in \mathbb{R}$, and one eigenvalue of $M$ is $2,$ then the largest among the absolute values of the eigenvalues of $M$ is _______

Comments

i got 5 as answer. someone confirm?

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the largest among the absolute values of the eigenvalues of M 

 

Does this highlighted part of the question mean,

Lets assume the Eigen values are 4,2, -5.

Then if we take absolute values of all these 3 values, the greatest value is 5, so the answer would be 5.

 

Someone pls confirm this .

 

Answer 1

so ans is 5

Answer 2

The characteristic polynomial $\lambda^3-4\lambda^2+a\lambda+30$ is three dimensional. So there will be 3 roots of this polynomial equation i.e. $\lambda^3-4\lambda^2+a\lambda+30=0 \tag{i}$

Putting $\lambda=2$ (as given in the question) to no$\mathrm{(i)}$, we get

$2^3-4\times2^2+2a+30=0\Rightarrow a=-11$.

So the characteristic polynomial equation is now $\lambda^3-4\lambda^2-11\lambda+30=0$. Since one of its roots is $2$, hence $(\lambda -2)$ will divide the polynomial.

So

$\begin{align}\lambda^3-4\lambda^2-11\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda^2-11\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda(\lambda-2)-15\lambda+30&=0\\\Rightarrow \lambda^2(\lambda-2)-2\lambda(\lambda-2)-15(\lambda-2)&=0\\\Rightarrow (\lambda-2)(\lambda^2-2\lambda-15)&=0\\ \therefore \lambda^2-2\lambda-15&=0\\\Rightarrow \lambda &= \frac{2\pm\sqrt{2^2+4\times15}}{2}\\ &=1\pm4 \\ &=5,-3 \end{align}$

 
$\therefore$ The Eigenvalues $\lambda=-3,2,5$.

So the largest absolute of $\lambda$ i.e. the maximum absolute Eigenvalue is $5$.

Answer 3

The characteristic equation of the 3*3 matrix

 

x^3 – tr(A)x^2 + (A11+A22+A33)x – det(A)=0  [ here lambda is same as x ]

 

A11 = cofactor(a11) ,  A11 = cofactor(a22) ,  A11 = cofactor(a33)

 

  now let's compare with the ch equation given in the question

 

we can see  :  tr(A) = 4 == > [ 2+x2+x3 = 4 ] == > x2+x3=2

                    : 2*x2*x3 = -30  == > x2*x3 = -15 == > x2=-3 and x3=5  (or) vice versa

 

they asked largest absolute one so answer is 5

Answer 4

As, λ=2  has been give so put in the eqn.

 λ^3 – 4λ^2 + aλ + 30 = 0  

8-16+2a+30 = 0

a = – 11

so ,  λ^3 – 4λ^2 + 11λ + 30 =0

(λ-2)(λ-5)(λ+3) = 0

λ = -3,2,5

So, 5 is the largest among themeselves.