Matrix Problem for year 2021!

Algebra Level pending

A A is an n × n n \times n matrix so that: A = [ 2 a 1 0 0 ⋯ 0 0 0 a 2 2 a 1 0 ⋯ 0 0 0 0 a 2 2 a 1 ⋯ 0 0 0 ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ 0 0 0 0 ⋯ a 2 2 a 1 0 0 0 0 ⋯ 0 a 2 2 a ] A=\begin{bmatrix} 2a & 1 & 0 & 0 & \cdots & 0 & 0 & 0\\ a^2 & 2a & 1 & 0 & \cdots & 0 & 0 & 0\\ 0 & a^2 & 2a & 1 & \cdots & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \cdots & a^2 & 2a & 1\\ 0 & 0 & 0 & 0 & \cdots & 0 & a^2 & 2a\\ \end{bmatrix}

Where a ≠ 0 a \neq 0 , X → = [ x 1 x 2 x 3 ⋯ x n ] T \overrightarrow{X}=\begin{bmatrix} x_1 & x_2 & x_3 & \cdots & x_n\end{bmatrix}^T , b → = [ 1 0 0 ⋯ 0 ] T \overrightarrow{b}=\begin{bmatrix} 1 & 0 & 0 & \cdots & 0\end{bmatrix}^T .

If A X → = b → A\overrightarrow{X}=\overrightarrow{b} , find 1 x 1 \dfrac{1}{x_1} when n = 2020 , a = 2020 n=2020, a=2020 .


The answer is 2021.

Alice Smith by Alice Smith , 20, China

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1 solution

Daniele Rosmondi
Dec 29, 2020

Start from the n n -th row:

a 2 x n − 1 + 2 a x n = 0 ⇒ x n = − 1 2 a x n − 1 a^2x_{n-1}+2ax_n=0\ \ \Rightarrow x_n=-\frac 12 ax_{n-1}

Now, consider the ( n − 1 ) (n-1) -th row and then substitute the expression obtained for x n 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 − 1 2 a x n − 1 = 0 ⇒ x n − 1 = − 2 3 a x n − 2 a^2x_{n-2}+2ax_{n-1}+x_n=0\\ a^2x_{n-2}+2ax_{n-1}-\frac 12 ax_{n-1}=0\\ \Rightarrow x_{n-1}=-\frac 23ax_{n-2}

Going on, or by induction on k k , we see that

x n − ( k − 1 ) = − k k + 1 a x n − k x_{n-(k-1)}=-\frac k{k+1} ax_{n-k}

In particular, for k = n − 1 k=n-1 ,

x 2 = − n − 1 n a x 1 x_{2}=-\frac {n-1} {n} ax_{1}

Finally, consider the 1st row and substitute the expression obtained for x 2 x_2 :

2 a x 1 + x 2 = 1 2 a x 1 − n − 1 n a x 1 = 1 ⇒ x 1 = n a ( n + 1 ) 2ax_1+x_2=1\\ 2ax_1-\frac {n-1} {n} ax_{1}=1\\ \Rightarrow x_1=\frac n{a(n+1)}

Substitute given values to get

x 1 = 2020 2020 ( 2020 + 1 ) = 1 2021 \displaystyle x_1=\frac {2020} {2020(2020+1)} =\frac 1{2021}

and so 1 x 1 = 2021 \frac 1{x_1}=2021 .

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