A is an n × n matrix so that: A = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 2 a a 2 0 ⋮ ⋮ ⋮ 0 0 1 2 a a 2 ⋮ ⋮ ⋮ 0 0 0 1 2 a ⋮ ⋮ ⋮ 0 0 0 0 1 ⋮ ⋮ ⋮ 0 0 ⋯ ⋯ ⋯ ⋱ ⋱ ⋱ ⋯ ⋯ 0 0 0 ⋮ ⋮ ⋮ a 2 0 0 0 0 ⋮ ⋮ ⋮ 2 a a 2 0 0 0 ⋮ ⋮ ⋮ 1 2 a ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
Where a = 0 , X = [ x 1 x 2 x 3 ⋯ x n ] T , b = [ 1 0 0 ⋯ 0 ] T .
If A X = b , find x 1 1 when n = 2 0 2 0 , a = 2 0 2 0 .
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Start from the n -th row:
a 2 x n − 1 + 2 a x n = 0 ⇒ x n = − 2 1 a x n − 1
Now, consider the ( n − 1 ) -th row and then substitute the expression obtained for x n :
a 2 x n − 2 + 2 a x n − 1 + x n = 0 a 2 x n − 2 + 2 a x n − 1 − 2 1 a x n − 1 = 0 ⇒ x n − 1 = − 3 2 a x n − 2
Going on, or by induction on k , we see that
x n − ( k − 1 ) = − k + 1 k a x n − k
In particular, for k = n − 1 ,
x 2 = − n n − 1 a x 1
Finally, consider the 1st row and substitute the expression obtained for x 2 :
2 a x 1 + x 2 = 1 2 a x 1 − n n − 1 a x 1 = 1 ⇒ x 1 = a ( n + 1 ) n
Substitute given values to get
x 1 = 2 0 2 0 ( 2 0 2 0 + 1 ) 2 0 2 0 = 2 0 2 1 1
and so x 1 1 = 2 0 2 1 .