in Linear Algebra
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in Linear Algebra
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3 Comments

(HERE X is the lambda)

as eigen vaues are given the characteristic equation must be having factors 2,2,-1

so characteristic equation after solving  | A-x*I |=0 will boil down to

(x-2)*(x-2)*(x+1)=0

expand this equation and put x=A(as every matrix can be represented as it's eigen value lambda)
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A.

(A+1)(A-2)$^{{2^{}}}$=0

Simplify the equation, multiply with A$^{-1}$, and put everything on RHS, except A$^{-1}$ on LHS. Answer will be  A.
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let check option by option,

option A:-

$P^{-1} = \frac{1}{4}(3P-P^2) $

$4.P^{-1} = (3P-P^2) $

multiply by P

$4I = (3P^2-P^3) $

this is the characteristic eqn, so substitute $\lambda$

$\lambda=2$, then 4 = 12-8 ===> satisfies

$\lambda=-1$, then 4 = 3-(-1) ===> satisfies

so A is correct option !
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