[1&1 0&1]-1[1&2 0&1]-1[1&3 0&1]-1....[1&100 0&1]-1=[1&a b&1], then absolute value of a+b is

Question

[1&1 0&1]-1[1&2 0&1]-1[1&3 0&1]-1....[1&100 0&1]-1=[1&a b&1], then absolute value of a+b is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

[1&-1 0&1][1&-2 0&1][1&-3 0&1]...[1&-100 0&1] =[1&-(1+2) 0&1][1&-3 0&1]...[1&-100 0&1] = [1&-(1+2+3) 0&1]...[1&-100 0&1] = [1&-(1+2+3+...+100) 0&1] ⇒ a=-(100(101)/2)=-5050, b=0 ∴ a+b=-5050

Q. $[1011]^{- 1} [1021]^{- 1} [1031]^{- 1} .... [101001]^{- 1} = [1 b a 1]$ , then absolute value of $a + b$ is

$[10 - 1 1] [10 - 2 1] [10 - 3 1] ... [10 - 100 1]$
$= [10 - (1 + 2) 1] [10 - 3 1] ... [10 - 100 1]$
$= [10 - (1 + 2 + 3) 1] ... [10 - 100 1]$
$= [10 - (1 + 2 + 3 + ... + 100) 1]$
$\Rightarrow a = - \frac{100 ( 101 )}{2} = - 5050, b = 0$
$∴ a + b = - 5050$

Q. [10​11​]−1[10​21​]−1[10​31​]−1....[10​1001​]−1=[1b​a1​], then absolute value of a+b is

[10​−11​][10​−21​][10​−31​]...[10​−1001​] =[10​−(1+2)1​][10​−31​]...[10​−1001​] =[10​−(1+2+3)1​]...[10​−1001​] =[10​−(1+2+3+...+100)1​] ⇒a=−2100(101)​=−5050,b=0 ∴a+b=−5050

Q. $[1011]^{- 1} [1021]^{- 1} [1031]^{- 1} .... [101001]^{- 1} = [1 b a 1]$ , then absolute value of $a + b$ is

$[10 - 1 1] [10 - 2 1] [10 - 3 1] ... [10 - 100 1]$
$= [10 - (1 + 2) 1] [10 - 3 1] ... [10 - 100 1]$
$= [10 - (1 + 2 + 3) 1] ... [10 - 100 1]$
$= [10 - (1 + 2 + 3 + ... + 100) 1]$
$\Rightarrow a = - \frac{100 ( 101 )}{2} = - 5050, b = 0$
$∴ a + b = - 5050$