CSIR NET MATHEMATICS | Linear Algebra | Eigenvalues of a Matrix DEC 2011 (हिंदी में)

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Oct 14, 2026
CSIR NET MATHEMATICS | Linear Algebra | Eigenvalues of a Matrix DEC 2011 (हिंदी में)
Linear Algebra, Eigenvalues of a Matrix,
CSIR NET Answer Key December 2011,
CSIR NET Answer Key June 2019, Mathematical Science,
Helpful those who are preparing for NET, GATE, SET, SLET exams
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Hello friends, in this video
Inside, we have Lenea LGprekevi
We will describe the problem.
CSN Night December 2011
The matrix A has certain complex numbers λ
not equal to 0 as an eigenvalue
Which of the following numbers must also be an eigenvalue of A?
We have 4 options; the first option is Lambda Plus 20.
second option lambda minus 20
third option 20 minus lambda
fourth option minus 20 minus lambda
This is a problem related to eigenvalues.
Then they will remember the results.
The first result is the sum of the eigenvalues of A.
equal to the trace of A
The product of the eigenvalues of A is equal to the determinant of A.
Both of these are very important results.
There are many problems that are used frequently.
We will first find the trace of A
trace of A means the sum of the diagonal entries
Here the diagonal values are 40, 30, and minus 50.
If we take their sum, the answer will be 20.
And the answer for determinant A is 0
A 3 by 3 matrix
Thus, there are 3 possible eigenvalues for it.
lambda 1, lambda 2, lambda 3
So the sum of the eigenvalues of A is the trace of A.
therefore
lambda 1 plus lambda 2 plus lambda 3
equal to the trace of A
20
The product of the eigenvalues of A is the determinant of A.
So lambda 1 into lambda 2 into lambda 3
equal to 0
The product of 3 numbers is 0.
I mean, at least one number should be 0
Here we are given that
A complex number is its eigenvalue, which is non-zero.
the conjugate of a complex number
It will also be its eigenvalue.
Then the second eigenvalue is its conjugate.
will be a complex number, and it will also be non-zero.
So two
the eigenvalues will become non-zero
Then there will be only one eigenvalue, which is 0.
That lambda 3 equals 0
lambda 1 plus lambda 2
plus lambda cubed equals 20
Lambda 3 is a species of life.
Then we will get λ₁ + λ₂
equal to 20
Let λ be an eigenvalue; suppose λ₁ = λ.
So we will get, lambda plus lambda 2 equals 20,
Then the value of lambda 2 will be 20 minus lambda.
Then the second value will be 20 minus lambda, if one value is lambda.
Option number 3 is correct.
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