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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 _______

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+5
i got 5 as answer. someone confirm?

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 $= 5$
by Boss (28.8k points)
selected by
+4
How to solve cubic equation quickly?
+38

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+5

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Nice and simple cubic explanation kajal
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could you please explain what's going on in the image you uploaded?

$a\lambda^3 – b \lambda^2 + c \lambda +d = \lambda^3 – 4 \lambda^2 + x \lambda +30$
(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.

by Veteran (102k points)
edited by
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Sum of eigen values   = sum of diagonal element
Sum of eigen values    =   (-b/a)
What is b ?

+7
$ax^3\ +\ bx^2\ +\ cx + d = 0\\ Let\ the\ roots\ be\ l,m,n\\l+m+n\ =\ -b/a\\l*m*n\ =\ -d/a$
$\lambda^3-4\lambda^2+a\lambda+30$

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$

$-\lambda^3+2\lambda^2$

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

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

$2\lambda^2-4\lambda$

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$-15\lambda+30$

$15\lambda-30$

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

Hence : Max $\{2,-3,5\}=\color{Red}5$
by Boss (11.9k points)
0
I followed this technique to find the answer in cube root
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
by Boss (18.1k points)

so ans is 5

by Loyal (6.8k points)
+1 vote
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 $5$.
by Active (3.6k points)