3

I tried doing this with regular method and computations became very complex .Is there any shorter way , Am i missing something ?

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Sophie Clad
  • 2,291

2 Answers2

3

Hint: Since all columns (or alternatively all rows) all equal, the rank of the matrix is 1

Droplet
  • 236
2

It can be done by using the characteristic equation:

Let us assume that the characteristic equaton of the matrix A is:

$$ \lambda^4 + a_1\lambda^3 + a_2\lambda^2 + a_3 \lambda + a_4 = 0$$

If you check then,

$$A^2 = 4 A$$

Using the fact that a matrix satisfies its own characteristic equation we get:

$$A^4 = 4 A^3 $$

Again writing it in terms of the Eigen values we get:

$$ \lambda^4 - 4 \lambda^3 = 0$$

Solve this and you will get two distinct Eigen values: 4 and 0.

FAM
  • 285