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How can I find the inverse of a n-by-n lower triangular matrix with diagonal elements all equal to 2, and lower triangular elements all equal to one?

\begin{bmatrix} 2&&&& \\ 1&2&&&\\ 1&1&2&&\\ \vdots& \ddots & \ddots & \ddots &\\ 1&\dots&1&1&2 \end{bmatrix}

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    $\begingroup$ What did you get for $n=2$ and $n=3$? $\endgroup$ – Dietrich Burde Jan 8 '18 at 19:07
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Try to work it out piecemeal, and then find a pattern and prove it in general.

What should the entry (1, 1) be? Logically, doing the first step of matrix multiplication with the matrix and its inverse (dot product of first row and first column), we see that the result has to be 1. The only way this can ever happen is if entry (1, 1) is $\frac12$.

What should all the other entries in the first row be? The result of dot products between the first row and all other columns of the inverse matrix should be 0, if we are to have the identity matrix as a result. This can only happen if all other such values are zero.

With this knowledge, proceed similarly for the next rows. You will have enough information to reason it out.

Now take this pattern and prove it for the general case.

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The inverse is again a lower triangular matrix, with $\frac{1}{2}$ on the main diagonal and $-\frac{1}{2^k}$ on the $k$-th subdiagonal. So the left lower corner is $-\frac{1}{2^n}$. Proof by computing $AA^{-1}=I_n$.

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Try a few small $n$, and you should see a pattern. Then prove it using induction.

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We have $$A^{-1}A=I\Leftrightarrow \begin{bmatrix}{C_1}&{C_2}&{\ldots}&C_n\end{bmatrix}A=\begin{bmatrix}{I_1}&{I_2}&{\ldots}&I_n\end{bmatrix}.$$ Now, solve the system and you'll get the columns of $A^{-1}$.

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Let $A$ be the matrix you specified, $I$ be the identity matrix, and $J := \begin{bmatrix} 0 & & & & \\ 1 & 0 & & & \\ 0 & 1 & \ddots & & \\ \vdots & \ddots & \ddots & 0 & \\ 0 & \cdots & 0 & 1 & 0 \end{bmatrix}$.

What we know is $J^n = 0$, so $A - I = \sum_{i=0}^\infty J^i$ and $(A-I)(I-J) = I$. Left multiply $A^{-1}$,

$$ (I - A^{-1})(I - J) = A^{-1} \quad \Rightarrow \quad I - A^{-1} - J + A^{-1}J = A^{-1}. $$

Hence

$$ \begin{aligned} A^{-1}(2I-J) &= I-J\\ A^{-1} &= \frac12(I-J)\left(I-\frac12J\right)^{-1} = \frac12(I-J)\sum_{i=0}^\infty \frac{J^i}{2^i} = \frac12\left(\sum_{i=0}^\infty \frac{J^i}{2^i} - \sum_{i=1}^\infty \frac{J^i}{2^{i-1}}\right), \end{aligned} $$

And remember that $J^n = 0$.

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