GATE CSE 2016 Set 1 | Question: 05

Question

Two eigenvalues of a $3 \times 3$ real matrix $P$ are $(2+\sqrt {-1})$ and $3$. The determinant of $P$ is _______

Answer 1

Let $A$ be a real symmetric matrix and let $\lambda$ be an eigen value and let $x$ be the corresponding eigen vector.

Now since $x$ is an eigen vector so we must have this relation:

$Ax$=$\lambda x$

$\overline {Ax} = \overline {\lambda x} $     (Taking conjugate on both side)

$\bar A \bar x = \bar \lambda \bar x $     (For any two matrices A and B we have $\overline {AB}=\bar A \bar B$)

$A \bar x = \lambda \bar x$     (Because A and $\lambda$ both are real so doing conjugation on them will not change them)

 

So from this we get that $\bar \lambda$ is also an eigen value of $A$ and the corresponding eigen vector is $\bar x$.

And if we apply this process on real $\lambda$ then it will be same but if we apply this on some complex $\lambda$ then

we will get a different eigen value.So applying conjugate on $2+i$ we get a new eigen value

and ,i.e, $2-i$.

So eigen values of P are : $3$ , $2+i$ and $2-i$.

Hence determinant of P = $3*(2+i)*(2-i)=15$.

Answer 2

Given the matrix P which is a 3x3 matrix so the characteristic equation | P-ßI |=0 must have 3 eigen values and for each eigen value we would have infinite eigen vectors corresponding to the eigen value.

Let us suppose the 1 remaining eigen value is ×. Thus the eigen values would be x, 2+i, 3 . Now we also know that if a system has a complex variable as a solution, then it implies that the conzugate of the complex variable must also exist. Hence x=2-i.

Thus the eigen values : 2-i, 2+i, 3.

Determinant : Product of the eigen values : (2-i)*(2+i)*3 = 15