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TIFR CSE 2025 | Part A | Question: 6

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2 2 votes

Consider the following matrix:
\[
A=\left[\begin{array}{lll}
3 & 0 & 1 \\
0 & 2 & 0 \\
x & 0 & 1
\end{array}\right]
\]
If all eigenvalues of the matrix are the same, what is the value of $x$ ?

  1. $2$
  2. $-2$
  3. $1$
  4. $-1$
  5. $0$

     

2 Answers

4 4 votes
Just Find the Determinant of the matrix it will be 6-2x and one by one substitute all the values of x then you will realize that at x=-1 then determinant is 8 so 8 can be written as 2*2*2 basic factors so yes. 2,2,2 is the eigen values and it is given in question eigen values are same so yes answer is -1 which is option D.
2 2 votes
Sum of Eigen values = Tr(A)  i.e, sum of Diagonal elements

So, Sum of Eigen values = 3 + 2 + 1 = 6

It is given that all the eigen value are same. So it is 2 , 2, 2

Now, Product of Eigen values = det ( A )

         2 × 2 × 2 = det (A)

                     8  =  6 - 2x

            So, x = -1

Hence, The correct answer is (d)  -1.
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