If A = [1&0 1&1] and A8 = aA +bI, then (a , b) =

Question

If A = [1&0 1&1] and A8 = aA +bI, then (a , b) =  (A) (8, 7) (B) (-7, 8) (C) (8, -7) (D) (-8, -7)

Tardigrade Admin · Accepted Answer

We have, A = [1&0 1&1] A2 = A . A = [1&0 1&1] [1&0 1&1] = [1&0 2&1] A3 = A2 . A = [1&0 2&1] [1&0 1&1] = [1&0 3&1] Similarly , A8 = [1&0 8 &1] Now, A8 = aA + bI ⇒ : :: [1&0 8 &1] = a [1&0 1&1] + b [1 &0 0 1] ⇒ : :: [1&0 8&1] = [ a+ b &0 a a+b ] ∴ : :: : a = 8 and a + b = 1 : : ⇒ : b = 1 - 8 = - 7

Q. If $A = [1101]$ and $A^{8} = a A + b I,$ then $(a, b) =$

$(8, 7)$

$(- 7, 8)$

$(8, - 7)$

$(- 8, - 7)$

Q. If A=[11​01​] and A8=aA+bI, then (a,b)=

(8,7)

(−7,8)

(8,−7)

(−8,−7)

We have, A=[11​01​] A2=A.A=[11​01​][11​01​]=[12​01​] A3=A2.A=[12​01​][11​01​]=[13​01​] Similarly , A8=[18​01​] Now, A8=aA+bI ⇒[18​01​]=a[11​01​]+b[10​01​] ⇒[18​01​]=[a+ba​0a+b​] ∴a=8 and a+b=1⇒b=1−8=−7

Q. If $A = [1101]$ and $A^{8} = a A + b I,$ then $(a, b) =$

$(8, 7)$

$(- 7, 8)$

$(8, - 7)$

$(- 8, - 7)$