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Two eigenvalues of a $3\times3$ real matrix $P$ are $(2+​ \sqrt-1)$ and $3$. The determinant of $P$ is ________.

  1. $0$
  2. $1$
  3. $15$
  4. $-1$
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The same question asked as NAT type in GATE CS 2016.

Reference: https://gateoverflow.in/39634/gate2016-1-05

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3 Answers

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4 votes
Eigen values always occur in conjugate pairs.So all the 3 roots will be 2+i,2-i,3. Determinant of P= product of eigen values=15

EDIT: It is wrong to say that “Eigen values always occur in conjugate pairs”. What makes them appear in conjugate pair is the fact that given matrix is a real matrix. Why?
It is because determinant of real matrix is a real number and the determinant is equal to product of all the eigen values.

$\lambda _{1}$$\lambda _{2}$$\lambda _{3}$$\lambda _{4}$….. = determinant

If determinant is real number only then the product $\lambda _{1}$$\lambda _{2}$$\lambda _{3}$$\lambda _{4}$… is forced to be a real number. This implies the number of imaginary eigen values should be even i.e. they must occur in pairs. Now why conjugate pairs? you can see that from this equation :- $\lambda$$=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

 

Therefore, since the matrix is real, the eigen values occur in pairs...and moreover in conjugate pairs. Hence third eigen value must be 2-i
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2 votes
2 votes
one eigen value=$2+\sqrt{-1}$=$2+i$

second eigen value=$2-i$

third eigenvalue=3

The determinant of P= Multipication of  eigen value

=$(2+i)*(2-i)*3$

=$5*3$

=$15$
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0 votes
A $3\times3$ matrix always has $3$ eigenvalues.The third eigenvalue is not mentioned. So the third eigenvalue is $0$. Now, the determinant of a matrix is the product of eigenvalues. So anything multiplied to zero is always zero. Therefore, option $a$ is correct.
Answer:

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