Matrix

Question

To find the product of the non-zero eigenvalues of the matrix is ____

$\begin{pmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \end{pmatrix}$

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I got ans -2

I solved like this

The product of the non-zero eigenvalues of the matrix is ____

$\begin{pmatrix} \left ( 1-\lambda \right ) & 0 & 0 & 0 & 1 \\ 0 &\left ( 1-\lambda \right ) & 1 & 1 & 0 \\ 0 & 1 & \left ( 1-\lambda \right )& 1 & 0 \\ 0 & 1 & 1 & \left ( 1-\lambda \right )& 0 \\ 1 & 0 & 0 & 0 & \left ( 1-\lambda \right ) \end{pmatrix}$

$\begin{pmatrix} \left ( 1-\lambda \right ) & 0 & 0 & 0 & 1 \\ 0 &\left ( 1-\lambda \right ) & 1 & 1 & 0 \\ 0 & 1 & \left ( 1-\lambda \right )& 1 & 0 \\ 0 & 1 & 1 & \left ( 1-\lambda \right )& 0 \\ \lambda& 0 & 0 & 0 & -\lambda \end{pmatrix}$  $R_{5}\leftarrow R_{5}-R_{1}$

$\begin{pmatrix} \left ( 1-\lambda \right ) & 0 & 0 & 0 & \left ( 2-\lambda \right )\\ 0 &\left ( 1-\lambda \right ) & 1 & 1 & 0 \\ 0 & 1 & \left ( 1-\lambda \right )& 1 & 0 \\ 0 & 1 & 1 & \left ( 1-\lambda \right )& 0 \\ \lambda& 0 & 0 & 0 & 0\end{pmatrix}$ $C_{5}\leftarrow C_{5}+C_{1}$

$\begin{pmatrix} \left ( 1-\lambda \right ) & 0 & 0 & 0 & \left ( 2-\lambda \right )\\ 0 &\left ( 1-\lambda \right ) & 1 & 1 & 0 \\ 0 & 1 & \left ( 1-\lambda \right )& 1 & 0 \\ 0 & 0 & - \lambda & \lambda  & 0 \\ \lambda& 0 & 0 & 0 & 0\end{pmatrix}$ $R_{4}\leftarrow R_{4}-R_{3}$

$=\lambda \begin{bmatrix} 0 &0 & 0 &\left ( 2-\lambda \right ) \\ \left ( 1-\lambda \right ) &1 &1 &0 \\ 1& \left ( 1-\lambda \right ) & 1 & 0\\ 0& \lambda &-\lambda & 0 \end{bmatrix}$

$=-\lambda \left ( 1-\lambda \right )\begin{bmatrix} \left ( 1-\lambda \right ) & 1 & 1\\ 1 & \left ( 1-\lambda \right ) &1 \\ 0 & \lambda & -\lambda \end{bmatrix}$

$=-\lambda \left ( 2-\lambda \right )\left [ \left ( 1-\lambda \right ) \left \{ \left ( 1-\lambda \right )\left ( -\lambda \right )-\lambda \right \}-1\left ( -\lambda \right )+1\left ( -\lambda \right )\right ]$

$=-\lambda \left ( 2-\lambda \right )\left [ \left ( 1-\lambda \right )\left ( \lambda ^{2}-2\lambda \right ) \right ]$

then $\lambda =-1,2,0$

Where is my mistake, plz tell me

Comments

 srestha For a matrix sum of the eigenvalues and trace of that matrix should be equal.

In your solution both are different, So there is something wrong with your approach.

Now you did one thing wrong here that you can't use both row and column operation. together.

Either you should use all column operation or all row operation. if you use both then the value of determinant could be changed.

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https://gateoverflow.in/2013/gate2014-2-47

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no I did wrong in calculation

ankit mentioned that