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A-2-0-1-2-h-and-k-are-numbers-so-that-A-2-hA-kI-where-I-1-0-0-1-find-the-value-of-h-and-k-




Question Number 32869 by Rio Mike last updated on 04/Apr/18
A= (((2       0)),((1       2)) ) ;h and k are numbers so  that A^2 =hA + kI,where I=  (((1        0)),((0        1)) ).  find the value of h and k.
A=(2012);handkarenumberssothatA2=hA+kI,whereI=(1001).findthevalueofhandk.
Commented by abdo mathsup last updated on 09/Apr/18
the csfactetistic  polynomial of A is  p_A (X) =det(A−XI)= determinant (((2−X      0)),((1           2−X)))  =(2−X)^2  = 4 −4X  +X^2    and kayley hsmilton  theorem give  p_A (A)=0 ⇒4I −4A +A^2 =0 ⇒  A^2  = 4A −4I  ⇒ h=4 and k=−4.
thecsfactetisticpolynomialofAispA(X)=det(AXI)=|2X012X|=(2X)2=44X+X2andkayleyhsmiltontheoremgivepA(A)=04I4A+A2=0A2=4A4Ih=4andk=4.
Answered by Joel578 last updated on 05/Apr/18
A^2  =  ((2,0),(1,2) )  ((2,0),(1,2) ) =  ((4,0),(4,4) )     ((4,0),(4,4) ) =  (((2h),(  0)),((  h),(2h)) ) +  ((k,0),(0,k) ) =  (((2h + k),(       0)),((      h),(2h + k)) )    h = 4  2h + k = 4 → k = −4
A2=(2012)(2012)=(4044)(4044)=(2h0h2h)+(k00k)=(2h+k0h2h+k)h=42h+k=4k=4

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