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

This may sound silly,but can’t we carry out elementary transformations and convert the matrix into upper triangular matrix? then the eigen values will be simply the diagonal elements...(R2 ← R2 + 2R2) ???

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Then it won't be the same system of linear equations.

Eigen values changes on elementary row or column transformation.

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@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

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

@shamim_ahmed how you are getting 2 & 5, 

characteristic equation will be 

let lambda = x

x²  -7x+6=0  if you solve this equation you will get (1,6)
or follow the above method ..!

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@ankit my bad. It was a silly calculation mistake. :)

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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 $1,1,1$

Answer 2

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

Answer 3

Let $\lambda$ be the eigen value.

then,

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

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

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

If we substitute the options then only Option $C.$ will satisfy.

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

Comments

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