in Linear Algebra
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Let $A$ be a $3× 3$ real matrix with all diagonal entries equal to $0$. If $1 + i$ is an eigenvalue of $A$, the determinant of $A$ equals

  1. $-4$
  2. $-2$
  3. $2$
  4. $4$
in Linear Algebra
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1 Answer

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Characteristic equation $|A-\lambda I|=0$

$\lambda_1=1+i \implies \lambda_2=1-i $

$\because$ all diagonal elements=0, $\Rightarrow \text{trace}=0$

Trace = sum of eigen values=0

$\implies\lambda_1+\lambda_2+\lambda_3=0$

$\implies (1+i)+(1-i)+\lambda_3=0$

$\implies \lambda_3=-2$


$|A|=\text{product of eigen values}=(1+i)(1-i)(-2)$

$\implies |A|=-4$
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2 Comments

how come 1-i another eigan value?
1
1
Complex conjugate
0
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