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+10 votes

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. $x = -4$, $y =10$
asked in Linear Algebra by Boss (18.3k points)
edited by | 1.2k views

3 Answers

+19 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.
answered by Boss (32.1k points)
edited by
+4 votes
Solve the equation 3x 2y=8 and x 2y=16 which i get x= -4 ,y = 10
answered by Boss (14.4k points)
–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.
answered by Boss (19.9k points)

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