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Consider the following matrix $$A = \left[\begin{array}{cc}2 & 3\\x & y \end{array}\right]$$ If the eigenvalues of A are $4$ and $8$, then

  1. $x = 4$, $y = 10$
  2. $x = 5$, $y = 8$
  3. $x = 3$, $y = 9$
  4. $x = -4$, $y =10$
in Linear Algebra
edited by
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2y-3x=32

x=-4 , b= 10 

3 Answers

26 votes
 
Best answer
Sum of eigenvalues is equal to trace (sum of diagonal elements) and product of eighen values is equal to the determinant of matrix

So, $2+y=8+4$ and  $2y-3x = 32$

Solving this we get $y = 10, x =-4.$

Option $D$ is answer.

edited by
4 votes
Solve the equation 3x 2y=8 and x 2y=16 which i get x= -4 ,y = 10
–2 votes
using one simple property...

The sum of eigen values is equal to the sum of the diagonal elements.

Given that the eigen values are 4 and 8, we have,

8+4 = y + 2

y = 10.

Now out of A and D, I don't find any difference in the options.
1 flag:
✌ Edit necessary (manish_pal_sunny)
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