GATE CSE 2010 | Question: 29

Question

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

• A. $x = 4$, $y = 10$
• B. $x = 5$, $y = 8$
• C. $x = 3$, $y = 9$
• D. $x = -4$, $y =10$

Comments

2y-3x=32

x=-4 , b= 10 

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Similar question: GATE CSE 2015 Set 1 | Question: 36

Answer 1

Sum of eigenvalues is equal to trace (sum of diagonal elements) and product of eigen 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.

Answer 2

Solve the equation 3x 2y=8 and x 2y=16 which i get x= -4 ,y = 10

Answer 3

I found one more interesting solution, which will give you intuition as per @sachin sir language.

Follow the steps:

1. Select Option 1, make a matrix. 
2. Subtract eigenvalues from the main diagonal.
3. Check if the matrix has linearly dependent columns.
4. Follow the above 3 steps for other options as well.

I know the solution provided by Pooja is faster but it is also an interesting way to think about the solution.

Comments

For both the eigen values $|A - \lambda I|$ must be linearly dependent.

Answer 4

As we know The product of eigenvalue is equal to Determinant. So in this question, eigenvalues are given 4 and 8. So are determinant is 32. Now we can put the value of X and Y and  check whether the determinant is equal to 32 or not

Answer 5

Answer: D 

• Sum of eigenvalues equals the Trace of the matrix, which is the sum of the main diagonal entries.

• Product of eigenvalues equals the Determinant of the matrix.

$$\begin{vmatrix}
 2 & 3 \\
x & y
\end{vmatrix}=2y-3x$$

$$2y-3x=\lambda_1.\lambda_2$$ $\text{ and }$

$$y+2=\lambda_1+\lambda_2$$

 

$$2y-3x=32$$ $\text{ and }$

$$y+2=12$$

 

$$\longrightarrow y=10\text{ and }x=-4$$