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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 _______
asked in Linear Algebra by Active (1.6k points)
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+5
i got 5 as answer. someone confirm?

5 Answers

+39 votes
Best answer
Given that $ \lambda = 2$ is an eigen value. So, it must satisfy characterstic equation.

$2^3 - 4*2^2 + 2a + 30 = 0 \Rightarrow \color{green}{a = -11}$

Characterstic eq : $\lambda^3 -4\lambda^2 - 11\lambda + 30$

$\Rightarrow (\lambda - 2)(\lambda - 5)(\lambda + 3) = 0$

$\lambda_1 = 2, \lambda_2 = 5$ and $\lambda_3 = -3$

Max Eigen Value $ = 5$
answered by Boss (28.3k points)
selected by
+2
How to solve cubic equation quickly?
+29

......

+1

..

0
Nice and simple cubic explanation kajal
0

could you please explain what's going on in the image you uploaded?

+28 votes

$a\lambda^3 – b \lambda^2 + c \lambda +d = \lambda^3 – 4 \lambda^2 + x \lambda +30$
(Using $x$ in place of $a$ on question to preserve the convention)

$\implies a = 1, b = -4, c = x, d = 30$

Sum of eigen values  $=   (-b/a)=   4.$

So,  $x_1  +  x_2  +  x_3    =   4$

Given $x_1 = 2$
So, $    x_2  +  x_3      =  4  -  2   =  2 \quad \to (1)$

Product of eigen values   $=  (-d/a) = -30.$

So,  $x_1 . x_2 . x_3       =   -30$
$\implies x_2 . x_3         =  -15 \quad \to (2)$

Solving $(1)$ and $(2)$, we get 

$x_2  =  -3   ,   x_3  =  5$

Hence, the absolute value of largest of 3 eigen values  =  5.

answered by Veteran (101k points)
edited by
0

Sum of eigen values   = sum of diagonal element 
Sum of eigen values    =   (-b/a)
What is b ?

+4
$ax^3\ +\ bx^2\ +\ cx + d = 0\\ Let\ the\ roots\ be\ l,m,n\\l+m+n\ =\ -b/a\\l*m*n\ =\ -d/a$
+5 votes
Given that λ=2λ=2 is an eigen value. So, it must satisfy characterstic equation.

23−4∗22+2a+30=0⇒a=−1123−4∗22+2a+30=0⇒a=−11

Characterstic eq : λ3−4λ2−11λ+30λ3−4λ2−11λ+30

⇒(λ−2)(λ−5)(λ+3)=0⇒(λ−2)(λ−5)(λ+3)=0

λ1=2,λ2=5λ1=2,λ2=5 and λ3=−3λ3=−3

Max Eigen Value =5
answered by Boss (17.9k points)
+4 votes
$\lambda^3-4\lambda^2+a\lambda+30$

Given $\lambda-2=0$

$\therefore a=-11$

Complete Equation : $\lambda^3-4\lambda^2-11\lambda+30$

To find other $2$ roots just divide the equation with $\lambda-2$ which is root itself :

       
$\lambda -2\Large)$ $\lambda^3-4\lambda^2-11\lambda+30\Large ($$\lambda^2-2\lambda-15$

           $-\lambda^3+2\lambda^2$

           --------------------

                    $-2\lambda^2-11\lambda$

                       $2\lambda^2-4\lambda$

                     --------------------

                                $-15\lambda+30$

                                   $15\lambda-30$

$\therefore $ $Eq^n=$ $(\lambda-2)( \lambda+3) (\lambda-5)$

Hence : Max $\{2,-3,5\}=\color{Red}5$
answered by Boss (11.7k points)
0
I followed this technique to find the answer in cube root
+2 votes

so ans is 5

answered by Loyal (6.7k points)
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