GATE CSE 2005 | Question: 49

Question

What are the eigenvalues of the following $2\times 2$ matrix? $$\left( \begin{array}{cc} 2 & -1\\ -4 & 5\end{array}\right)$$

• A. $-1$ and $1$
• B. $1$ and $6$
• C. $2$ and $5$
• D. $4$ and $-1$

Comments

@preeti0448 then the eigen values will change because eigen value changes  for elementary row or column transformations.

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when we are doing elementary row operation then we are getting diagonal elements as 2,3..which will be eigen value...but it is not in options..now when we do it by characterstics equation we get eigen value as 1,6.which is given in option...my doubt is if we have both of them in options then which one will be true?? @Sachin Mittal 1

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@Aman Dabral  

The question will be incorrect then, as we can't perform elementary row operations for finding eigenvalues.

Answer 1

Let the eigen values be $a,b$

Sum of Eigen Values = Trace(Diagonal Sum)

 $\implies a+b = 2+5 = 7$

Product of Eigen Values = Det(A)  

 $\implies a\times b = 6$

Solving these we get eigenvalues as 1 and 6. 

Option(B) is Correct.

Comments

Important properties of Eigen values:-

$(1)$Sum of all eigen values$=$Sum of leading diagonal(principle diagonal) elements=Trace of the matrix.

$(2)$ Product of all Eigen values$=Det(A)=|A|$

$(3)$ Any square diagonal(lower triangular or upper triangular) matrix eigen values are leading diagonal (principle diagonal)elements itself.

 

Example$:$$A=\begin{bmatrix} 1& 0& 0\\ 0&1 &0 \\ 0& 0& 1\end{bmatrix}$

    Diagonal matrix

  Eigenvalues are $1,1,1$

$B=\begin{bmatrix} 1& 9& 6\\ 0&1 &12 \\ 0& 0& 1\end{bmatrix}$

Upper triangular matrix

  Eigenvalues are $1,1,1$

$C=\begin{bmatrix} 1& 0& 0\\ 8&1 &0 \\ 2& 3& 1\end{bmatrix}$

Lower triangular matrix

  Eigenvalues are $1,1,1$

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Best property of eigen values

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@Lakshman Bhaiya 

In GoPDF Vol1 -> This Question (6.3.3 on pg 119) has misprint in matrix value of a12 [instead of showing it -1 , it is given as +1)

 

cc: Arjun sir (unable to tag him since typing @ arjun sir never give his name as option, even after adding "+" Suresh as well) , @gatecse 

Answer 2

Let $\lambda$ be the eigen value.

then,

$\begin{vmatrix} 2- \lambda &-1 \\ -4 & 5- \lambda \end{vmatrix}=0$

$\implies \lambda^2 -7 \lambda +10 -4 = 0$

$\implies \lambda^2 -7 \lambda +6 = 0$

$\implies \lambda^2 -6 \lambda -\lambda+6 = 0$

$\implies (\lambda -6)( \lambda -1) = 0$

$\implies \lambda = 6$  or  $\lambda = 1$

$\therefore$ Option $B$ is the correct answer.

Comments

no, option B (1,6) is correct. Please edit. It may confuse readers.

Answer 3

(2-x)(5-x)-4=0 x=1,6

Answer 4

Sum of eigenvalues = trace = 2+5=7

Product of eigenvalues = determinant = 2⋅5 − (−1)(−4) = 10 − 4 = 6

Only pair satisfying sum = 7 and product = 6 is:

(1,6)

 

Answer 5

Correct answer is - B , look at these 2 Method