Solving a system of linear equations using matrices

Question

I am trying to find a solution to the equation $\pi A = 0$ where $A$ is an $n\times n$ matrix as follows:

$$A = \begin{bmatrix} -6 & 6 & 0 \cdots & \, & \, & \, & 0\\ 1 & -7 & 6 & 0 &\cdots & \, & 0 \\ 0 & 2 & -8 & 6 & 0 &\cdots & 0 \\ \vdots & \ddots & \\ 0 & \dots & \dots & \, && n-1 & -n-5 \end{bmatrix}$$

So I have $-k-5$ as the diagonal element for the $k^\text{th}$ row, $6$ above the diagonal and $k$ as the element below the diagonal element for the $k^\text{th}$ row. May I know how I can solve for $\pi$ which is a row vector of size $n$?

Answer 1

In general, if $A$ is a $n\times n$ matrix, the system $xA = 0$ has an unique solution if and only if $A$ is row-equivalent to the $n\times n$ identity matrix $I_{n\times n}$.

Now, the matrix $A$ you presented in the problem is indeed row-equivalent to $I_{n\times n}$. I will not provide a full proof, but a sketch instead, which I hope to be useful.

I came with this solution by manually row-reducing $A$ from $n=1$ up to $n=4$ and noticing I was actually doing the same steps over and over. I'll show you what I did when row-reducing the $n=4$ case in order; first let's get $A$ to echelon form:

1. Multiply row $1$ by $-1/6$, then add $-1$ times row $1$ to row $2$.
2. Then, multiply row $2$ by $-1/6$, then add $-2$ times row $2$ to row $3$.
3. Then, multiply row $3$ by $-1/6$, then add $-3$ times row $3$ to row $4$.
4. Then, multiply row $4$ by $-1/6$.

Now, let's get $A$ to it's row-reduced-echelon form:

1. Add $1$ times row $4$ to row $3$.
2. Add $1$ times row $3$ to row $2$.
3. Add $1$ times row $2$ to row $1$.

In general, for any $n$, if you want to get from $A$ to $I_{n\times n}$, the steps to take are:

1. To get $A$ in echelon form: for each $ 1 \leq i \leq n$, starting from $i=1$, first multiply the $i$-th row by $-1/6$, then add $-i$ times row $i$ to row $(i+1)$ (except when $i=n$, because there's no $(n+1)$-th row in a $n\times n$ matrix).
2. To reduce it: for each $1 \leq i \leq n$, add $1$ times row $i$ to row $(i-1$) (again, except when $i=1$, for there is no $0$-th row in a matrix).
3. $A$ is now in row reduced echelon form, and it is row-equivalent to $I_{n\times n}$.

This means that the system $xA=0$ has indeed an unique solution, which is $x=0$.