2 votes 2 votes Linear Algebra zeal linear-algebra discrete-mathematics matrix zeal2019 + – Prince Sindhiya asked Nov 17, 2018 edited Mar 9, 2019 by ajaysoni1924 Prince Sindhiya 359 views answer comment Share Follow See all 3 Comments See all 3 3 Comments reply Lakshman Bhaiya commented Nov 17, 2018 i edited by Lakshman Bhaiya Nov 17, 2018 reply Follow Share $(1)$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. 0 votes 0 votes Prince Sindhiya commented Nov 18, 2018 reply Follow Share @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 0 votes 0 votes Lakshman Bhaiya commented Nov 18, 2018 reply Follow Share Yes,see this https://www.quora.com/What-is-the-difference-between-complex-conjugates-and-irrational-conjugates 0 votes 0 votes Please log in or register to add a comment.