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Consider the following matrix:

$\left [ A \right ]=$ $\begin{bmatrix} 1.5&0 &1 \\ -0.5 & 0.5& -0.5\\ -0.5& 0 & 0 \end{bmatrix}$

The eigen values of the above matrix are:

  1. $-0.5,  0.5, 1.5$
  2. $0.5, 0.5, 1$
  3. $1.75, -1, 0.5$
  4. $-0.5,  0.5, -1$
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2 Answers

Best answer
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4 votes

The determinant of above matrix is 0.25.

Using the property, "The product of eigen values of any matrix is equal to its determinant."

Let's take the eigen values (0.5,0.5,1), then its product is 0.25 which is equal to the determinant of the matrix.

Hence the answer is option b. 

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  • The trace of A, defined as the sum of its diagonal elements, is also the sum of all eigenvalues,

{\displaystyle \operatorname {tr} (A)=\sum _{i=1}^{n}A_{ii}=\sum _{i=1}^{n}\lambda _{i}=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{n}.}

only option B

Answer:

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