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Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2+\sqrt {-1})$ and $3$. The determinant of $P$ is _______
asked in Linear Algebra by Boss (8.8k points)   | 883 views

2 Answers

+23 votes
Best answer
Given two eigen values are (2+i) and 3.. since it is a real matrix the 3rd eigen value is 2-i
determinant of P = product of eigen values.
Solving we get,

Answer 15.
answered by Boss (8.6k points)  
edited by
no1 is telling if the answers are correct or not.atlst upvotes.. :(
sure u r right .. but the 3rd eigen value is 2 - i because is is 3x3 matrix nt because it is real matrix .
and the complex root are in pair + and -
+4 votes
Eigen values are roots of Characterstic equation $|A - \lambda I | = 0$

For a $3×3$ matrix, characterstic equation will be cubic, so will have $3$ roots. Two roots are given as: $ 2 + i$ and $3$ and We know that complex roots always occur in pairs so, if $2+i$ is a root of characterstic equation, then $2-i$ must be other root.

$\lambda_{1} = 2+i$, $\lambda_{2} = 2-i$ and $\lambda_{3} = 3$

$\color{blue}{\det(A) = \lambda_{1}\lambda_{2}\lambda_{3} = (2+i)*(2-i)*3 = (2^2 - i^2)*3 = 5*3 = 15}$
answered by Veteran (22k points)  
edited by
we "can have". Not "will have" :)
Why Can ?
So, it's not compulsory?
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