Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1]],` the value of ` a + b + c + d ` is

Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1

1.	Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1]],` the value of ` a + b + c + d ` is
Answer» Correct Answer - 6 `because A = [[0 ,alpha],[0,0]]` `therefore A^(2) = Acdot A = [[0 ,alpha],[0,0]][[0 ,alpha],[0,0]]=[[0 ,0],[0,0]]=0` `rArr A^(2) = A^(3) = A^(4) = A^(5) = ...= 0` Now, `(A + I) ^(70) = (I+A)^(70)` `= I + ""^(70)C_(1) A + ""^(70)C_(2) A^(2) + ""^(70)C_(3) A^(3) +...+ ""^(70)C _(70)A^(70` `= I + 70 A + 0 + 0 + ...=I+70A` `rArr (A+I) ^(70) - 70 A = I = [[1,0],[0,1]]= [[a-1,b-1],[c-1, d-1]]` [given ] `therefore a- 1 = 1, b-1 = 0, c-1 = 0, d-1=1` `rArr a=2, b=1, c=, d=2` Hence, `a + b + c + d = 6`

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Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1]],` the value of ` a + b + c + d ` is

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