NIELIT 2016 MAR Scientist C - Section B: 8

Question

The eigenvalues of the matrix $\begin{bmatrix}1 & 2\\ 4 & 3 \end{bmatrix}$ are

• A. $\text{5 and -5}$
• B. $\text{5 and -1}$
• C. $\text{1 and -5}$
• D. $\text{2 and 3}$

Answer 1

It can easily be solved using the property of Eigen Values :

1. The sum of the Eigen Values of a Matrix is equal to the Trace of the Matrix(Trace of the matrix is the sum of its leading diagonal).

2. The Product of the Eigen Values is Equal to the Determinant of matrix .

Here Determinant of Matrix is -5 so by using property 2 option A and D eliminated.

now by using property 1 option C eliminated so option B is the answer.

 

 

Answer 2

After solving the quadratic equation we will get eigen value 5 and -1. 

Comments

Can you please tell me how did you subtracted 8 from (1-λ)(3-λ) ??
I get it that you subtracted the product of 2 and 4,but please explain me the logic behind it.
Thanks

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We are trying to solve | A – λI | = 0,  here A is matrix , λ is eigen values and I is identity matrix and two parallel lines | | represent determinant.

Answer 3

To find Eigen values of 2x2 matrix [a11 a12] [b21 b22] we can use the formula called as

x2 - Trace(A) + |A|

Trace(A) = a11 + b22
|A| = (a11)(b22) - (b21)(a12)

after solving that quadratic equation the values of x would be the eigen values of that 2x2 matrix.

[1 2]
[4 3]

Trace(A) = 1 + 3 = 4
|A| = (1)(3) - (2)(4) = 3 - 8 = -5

x^2 - 4x - 5
x^2 + 1x - 5x - 5
x(x + 1) - 5 (x + 1) = (x - 5) (x + 1)
x = 5, -1

So, 5 and -1 are the eigen values of the above matrix.