Zeal Test Series 2019: Linear Algebra - Matrices

Question

 

Comments

$(1)$Trace of the matrix $A=$Sum of the leading diagonal element (main diagonal sometimes principal diagonal, primary diagonal)$=$Sum of all Eigen-values.

$(2)$Product of all Eigen values$=Det(A)=|A|$

Suppose $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$

$(1)$Trace of the matrix$= \lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}+\lambda_{5}$

$(2) \lambda_{1}.\lambda_{2}.\lambda_{3}.\lambda_{4}.\lambda_{5}=|A|$

In given question$:$

In case singular matrix $|A|_{5\times5}=0$

$\bullet$The complex conjugate of $a+ib$ is $a-ib$ where $i=\sqrt{-1}$
$\bullet$The conjugate of  $a+b\sqrt{n}$ is  $a-b\sqrt{n}$

Given that 

$\lambda_{1}=2+3\sqrt{-2}=2+3\sqrt{-1}.\sqrt{2}=2+3\sqrt{2}i$

$\lambda_{2}=2-3\sqrt{2}i$

$\lambda_{3}=2-3\sqrt{2}$

$\lambda_{4}=2+3\sqrt{2}$

$\lambda_{5}=0$

For many non trivial solutions $AX=0 (|A|=0)$

So$,(1)$ and $(3)$ are right.

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@lakshman here only two eigen values are given then you said

∙∙The complex conjugate of a+ib is a−ib
∙∙how you got this that if we have one eigen vale in complex form then other will be conjugate of this

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Yes,see this https://www.quora.com/What-is-the-difference-between-complex-conjugates-and-irrational-conjugates