Test by Bikram | Mock GATE | Test 1 | Question: 39

Question

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:

• A. $-0.5,  0.5, 1.5$
• B. $0.5, 0.5, 1$
• C. $1.75, -1, 0.5$
• D. $-0.5,  0.5, -1$

Answer 1

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. 

Comments

sir ,
could you solve this sum
and
are we allowed to apply elementary row operations ( eg: multiplying -2 to row 2 ) and then apply |A- lambda*Identity| = 0

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@harsh181996 i don't think that elementary row operations can be applied. i applied row operations and got option d as answer but then i searched and found that row operations changes the eigen values... so i think they should be solved by general eigen value finding method.

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sum of eigen values= trace of matrix

option d eliminited

Answer 2

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

only option B