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$$\begin{array}{l} \left( \begin{array}{l} \begin{array}{*{20}{c}} 1 & 2 & 3 & \cdots & n \\ \end{array} \\ \begin{array}{*{20}{c}} 2 & 3 & 4 & \cdots & {n + 1} \\ \end{array} \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\ \begin{array}{*{20}{c}} n & n+1 & n+2 & \cdots & {2n - 1} \\ \end{array} \\ \end{array} \right)\left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{x_2}} \\ \vdots \\ {{x_n}} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_1}} \\ {{b_2}} \\ \vdots \\ {{b_n}} \\ \end{array}} \right) \\ \left( \begin{array}{l} \begin{array}{*{20}{c}} 1 & 2 & 3 & \cdots & n \\ \end{array} \\ \begin{array}{*{20}{c}} 1 & 1 & 1 & \cdots & 1 \\ \end{array} \\ \begin{array}{*{20}{c}} \vdots & \vdots & \vdots & \cdots & \vdots \\ \end{array} \\ \begin{array}{*{20}{c}} 1 & 1 & 1 & \cdots & 1 \\ \end{array} \\ \end{array} \right)\left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{x_2}} \\ \vdots \\ {{x_n}} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_1}} \\ {{b_2} - {b_1}} \\ \vdots \\ {{b_n} - {b_{n - 1}}} \\ \end{array}} \right) \\ \end{array}$$

I started reducing the matrix; Am I on the right path? How should I continue?

EDIT:

current state is:

$$ \left( \begin{array}{l} \begin{array}{*{20}{c}} 1 & 2 & 3 & \cdots & n \\ \end{array} \\ \begin{array}{*{20}{c}} 1 & 1 & 1 & \cdots & 1 \\ \end{array} \\ \begin{array}{*{20}{c}} 0 & 0 & 0 & \cdots & 0 \\ \end{array} \\ \begin{array}{*{20}{c}} \vdots & \vdots & \vdots & \cdots & \vdots \\ \end{array} \\ \begin{array}{*{20}{c}} 0 & 0 & 0 & \cdots & 0 \\ \end{array} \\ \end{array} \right)\left( {\begin{array}{*{20}{c}} {{x_1}} \\ {{x_2}} \\ \vdots \\ {{x_n}} \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_1}} \\ {{b_2} - {b_1}} \\ \begin{array}{l} {b_3} - 2{b_2} + {b_1} \\ \vdots \\ \end{array} \\ {{b_n} - 2{b_{n - 1}} + {b_{n - 2}}} \\ \end{array}} \right) $$

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2  
Yes, you then can make all rows except first and second rows are zero – Jlamprong Jan 19 '14 at 17:43
1  
Are the bottom-left entries $2,3$ wrong? – pppqqq Jan 19 '14 at 17:47
    
@pppqqq, thanks. it's a typo – SuperStamp Jan 19 '14 at 17:51
    
@Jlamprong, Thanks, but then I get stuck with $n-2$ zero rows with the form of $0...0=b_n-2b_{n-1}+b_{n-2}$. How to treat them? – SuperStamp Jan 19 '14 at 18:02
    
If $b_{k+2}-2b_{k+1}+b_k \ne 0$ for any $k \ge 1$, then the system has no solutions. Otherwise, you get $n-2$ equations of the form $0_{1 \times n} x = 0$, which are always true and can be discarded, so your system is reduced to the one defined by the first two rows of $A$ and $b$ (2 equations, $n$ unknowns). – Vedran Šego Jan 6 at 11:26

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