# Calculating the Inverse of a matrix.

$A)$

Solve:

$x_1-x_2=1$

$-x_1+2x_2-x_3=0$

$\vdots$

$-x_{99}+2x_{100}=0$

$B$)Deduce the inverse of

A=\begin{pmatrix} 1 & -1 & 0 & \cdots& \cdots & 0 \\ -1 & 2 & -1 & \ddots& \cdots &\vdots\\ 0 & -1 & 2 &-1& \ddots& \vdots\\ \vdots & \ddots & \ddots & \ddots &\ddots&0\\ \vdots & \cdots & \ddots & \ddots &2&-1\\ 0 & \cdots & \cdots & 0 &-1&2\\ \end{pmatrix}

I've solved the first part and got:

$x_1=100$

$x_2=99$

$\vdots$

$x_{100}=1$ (Correct me if I'm wrong)

For the part B , I'm not sure how to approach it although I'm quite sure the first column of $A^{-1}$ should contain $x_1$ till $x_{100}$ in this order , however I have no idea on how to fill the other columns.

If anyone could help me or give me hints , I would be grateful.

• Is the first entry supposed to be a $2$? I'm not sure what the pattern is. – Elliot G Feb 22 '18 at 19:23
• are you sure about the 1 in the first row and first column in your matrix. – Surb Feb 22 '18 at 19:35
• Yes I'm pretty sure since if you multiply the matrix A by the column matrix containing $x_1$ to $x_{100}$ you'd get the system given in part $A)$ – Raku Feb 22 '18 at 19:42

Here is a "brute-force" method.

Looking at some small dimension examples, it is clear how to construct the inverse. It is good to notice that $A$ is symmetric, and so its inverse is also symmetric.

Let $$B=\begin{bmatrix} 100&99&98&97&\cdots&2&1\\ 99&99&98&97&\cdots&2&1\\ 98&98&98&97&\cdots&2&1\\ 97&97&97&97&\cdots&2&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 1&1&1&1&\cdots&1 \end{bmatrix}$$ Now, when we do $BA$, we need to look at the three kinds of columns $A$ has:

• against the first colunm: $$\begin{bmatrix} 100&99\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=1,$$ and $$\begin{bmatrix} m&m\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=0,$$ so $(BA)_{k1}=\delta_{k1}$.

• against columns $2$ to $99$: the $j^{\rm th}$ column of $A$ has $-1,2,-1$ starting at row $j-1$. The $k^{\rm th}$ row of $B$ has $101-k$ in the first $k$ columns, and then starts decreasing one by one. So, for $j> k$, $$(BA)_{kj}=\begin{bmatrix}101-k&101-k-1&101-k-2\end{bmatrix}\begin{bmatrix}-1\\2\\-1\end{bmatrix}=0.$$ For $j<k$, the entry is the same by symmetry. For $j=k$, $$(BA)_{kk}=\begin{bmatrix} 101-k&101-k&101-k-1\end{bmatrix}\begin{bmatrix}-1\\2\\-1\end{bmatrix} =1$$ So $(BA)_{kj}=\delta_{kj}$.

• Against column $100$: when $k<100$, $j=100$, $$(BA)_{kj}=\begin{bmatrix} 2&1\end{bmatrix}\begin{bmatrix} -1\\2\end{bmatrix}=0.$$ And $$(BA)_{100,100}=\begin{bmatrix} 1&1\end{bmatrix}\begin{bmatrix} -1\\2\end{bmatrix}=1.$$ So $(BA)_{k,100}=\delta_{k,100}$.

In summary, $BA=I$.

Realize that what you calculated is a solution to $Ax=e_1$. If you are curious about a solution for $e_2$ set the first equation to 0, i.e. $x_1-x_2=0$ and the second to 1, yielding the same system of equation just one less variable (cause $x_1=x_2$). Then progress similarly and when you (conceptually) solve all, you have the solutions $A\mathbf{x}_i=e_i$.

I agree with you and your calculations of $\{x_1,\cdots, x_{100}\}$

And the first column of $A^{-1}$ is $\begin{bmatrix} x_1\\ \vdots\\x_{100} \end{bmatrix}$

The second column, the every row (except the first) will multiply by $\begin{bmatrix} 1\\-1\\0 \\\vdots\\0 \end{bmatrix}$ and result in $0.$

The second column is, $\begin{bmatrix} x_1-1\\x_2\\x_3\\ \vdots\\x_{100} \end{bmatrix}$

And from extending along I get:

$\begin{bmatrix} 100&99&98 &\cdots& 1\\ 99&99&98 &\cdots& 1\\ 98&98&98&\cdots&1\\ \vdots&&\ddots&&\vdots\\ 1&&\cdots&&1\end{bmatrix}$

$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}^{-1}=\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$

$\begin{pmatrix} 1 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 2 \end{pmatrix}^{-1}=\begin{pmatrix} 3 & 2 & 1\\ 2 & 2 & 1\\ 1 & 1 & 1 \end{pmatrix}$

$\begin{pmatrix} 1 & -1 & 0 & 0\\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2 \end{pmatrix}^{-1}=\begin{pmatrix} 4 & 3 & 2 & 1\\ 3 & 3 & 2 & 1\\ 2 & 2 & 2 & 1\\ 1 & 1 & 1 & 1 \end{pmatrix}$

Can you guess the inverse of $A$?