Question

If A is non-singular and [A-2L] [A - 4L] =0
then find [1/6] A + [4/3] [1/A] = ?
​

Answer 1

(A-2I)(A-4I) =0

A^2–4A-2A+8I=0

A^2–6A+8I=0

A^2–6A=-8I

Now, multiplying both sides of the above equation

we get

(A^-1)(A^2–6A) = A^-1 (-8I)

A-6I=-8(A^-1)

Divide both sides by 6, we get

(1/6) A-I=(-4/3)A^-1

(1/6)A+(4/3) A^-1= I

Hence, option b) I is correct answer.

I hope you understand my answer