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Consider the $5\times 5$ matrix:

$\begin{bmatrix} 1 & 2 &3 & 4 &5 \\ 5 &1 &2 & 3 &4 \\ 4& 5 &1 &2 &3 \\ 3& 4 & 5 & 1 &2 \\ 2&3 & 4 & 5 & 1 \end{bmatrix}$

It is given $A$ has only one real eigen value. Then the real eigen value of $A$ is ________
closed as a duplicate of: Gate ECE 2017 Eigen Value
in Linear Algebra by Veteran (117k points)
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15 will be the answer.

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$\left | A - \lambda I \right | =0$

$\Rightarrow \begin{vmatrix} 1 - \lambda & 2 & 3 & 4 & 5 \\ 5 & 1 - \lambda & 2 & 3 & 4 \\ 4 & 5 & 1 - \lambda & 2 & 3\\  3 & 4 &5 & 1 - \lambda &2 \\ 2 & 3 & 4 & 5 & 1 - \lambda \end{vmatrix} = 0$

Perform the row operation: $ R_1 \rightarrow R_1 + R_2 + R_3 + R_4 + R_5 $

$\Rightarrow \begin{vmatrix}
15 - \lambda & 15 - \lambda & 15 - \lambda & 15 - \lambda & 15 - \lambda \\
5 & 1 - \lambda & 2 & 3 & 4 \\
4 & 5 & 1 - \lambda & 2 & 3\\
3 & 4 &5  & 1 - \lambda &2 \\
2 & 3 & 4 & 5 & 1 - \lambda
\end{vmatrix} = 0 $

So, $ \lambda = 15 $ is the eigen value.
by (235 points)
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$\begin{bmatrix} 1 & 2 &3 & 4 &5 \\ 5 &1 &2 & 3 &4 \\ 4& 5 &1 &2 &3 \\ 3& 4 & 5 & 1 &2 \\ 2&3 & 4 & 5 & 1 \end{bmatrix}$

The characteristic equation is

$\begin{vmatrix} A-\lambda I \end{vmatrix}=0$

$\Rightarrow$$\begin{vmatrix} 1-\lambda & 2 &3 & 4 &5 \\ 5 &1-\lambda &2 & 3 &4 \\ 4& 5 &1-\lambda &2 &3 \\ 3& 4 & 5 & 1-\lambda &2 \\ 2&3 & 4 & 5 & 1-\lambda \end{vmatrix}$ =0

Applying $C_{1}\leftarrow C_{1}+C_{2}+C_{3}+C_{4}+C_{5}$

$\Rightarrow$$\begin{vmatrix} 15-\lambda & 2 &3 & 4 &5 \\ 15-\lambda &1-\lambda &2 & 3 &4 \\ 15-\lambda & 5 &1-\lambda &2 &3 \\ 15-\lambda& 4 & 5 & 1-\lambda &2 \\ 15-\lambda&3 & 4 & 5 & 1-\lambda \end{vmatrix}$ =0

Taking $15-\lambda$ as common

$\Rightarrow$$(15-\lambda)$$\begin{vmatrix} 1 & 2 &3 & 4 &5 \\ 1 &1-\lambda &2 & 3 &4 \\ 1& 5 &1-\lambda &2 &3 \\ 1& 4 & 5 & 1-\lambda &2 \\ 1&3 & 4 & 5 & 1-\lambda \end{vmatrix}$=0

$\Rightarrow$$(15-\lambda)=0$

$\Rightarrow$$\lambda=15$
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