0
$\begingroup$

So, $A$ is a $n$ x $n$ matrix where:

  • All the diagonal elements are $1$
  • All the elements on the lower triangular part are $-1$
  • The elements on the $n^{th}$ column are $1$
  • And every other element is $zero$

I am trying to find $b$ such that we already know what $x$ is without caring about the value of $n$.

I tried to use a $b$ similar to the one we use in $H_n x=b$ where $H_n$ is a $n$ x $n$ Hilbert matrix but still I couldn't specifically define the $x$.

Any hint would be appreciated!

  • 0
    What do you mean by elements of $n^{th}$ – 2017-02-02
  • 0
    the elements on the $n_{th}$ column, I corrected it. – 2017-02-02

1 Answers 1

1

I suggest that you try $b$ = first column of $A$; then $x = e_1$. Alternatively, pick $b$ = last column of $A$ and then $x = e_n$. Indeed, this works for any column. :)

  • 0
    I am sorry but what $e_n$ is? – 2017-02-02
  • 0
    $e_k$ is a vector of all zeroes except the $k$th entry, which is a $1$. In $\Bbb R^3$, for instance, $e_2$ is $\begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}$ and $e_3$ is $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$. For any dimension $k$, the vector $e_k$ is all zeroes except the last element, which is $1$. – 2017-02-02
  • 0
    Oh, that seems right. Thank you! – 2017-02-02