inverse of a tridiagonal matrix

Let $${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}} {-2}&{1}&{}&{}&{}\\ {1}&{-2}&{1}&{}&{}\\ {}&{1}&{\ddots}&{\ddots}&{}\\ {}&{}&{\ddots}&{\ddots}&{1}\\ {}&{}&{}&{1}&{-2} \end{array}} \right]$$ be a tridiagonal matrix. How we can prove that its inverse is the matrix $B=(b_{ij})$ where $$b_{ij}=-\frac{i(n+1-j)}{n+1} \; ,\quad i\leq j.$$

• Compute each entry of $AB$ explicitly? – hmakholm left over Monica Jan 2 '15 at 16:27
• And $b_{ij} = 0$ for $i>j$? – user14717 Jan 2 '15 at 16:43
• @NickThompson $A$ is symmetric so $B$ is as well ($b_{ij}=b_{ji}$ for $i>j$). – Algebraic Pavel Jan 2 '15 at 16:54
• @MathMan: By the rule for matrix multiplication, each entry of $AB$ is the sum of no more than three terms. You can write down explicit formulas for them and simplify. – hmakholm left over Monica Jan 2 '15 at 17:32
• It is quite straightforward (but rather tedious) to derive this formula using the fact that $\det A_{n\times n}=(-1)^n(n+1)$ (which follows from this recursion) and the Cramer's rule by expanding the minors by the suitable rows/columns. – Algebraic Pavel Jan 2 '15 at 17:46

As Henning points out, checking that two matrices are inverses is much easier than computing an inverse; all we need to do is find the product of the matrices. Also, note that your definition of $B$ is incomplete, but we can deduce the rest since $B$ must be symmetric.
By definition, the $i,j$ of the matrix product $AB$ is given by $$[AB](i,j) = \sum_{k=1}^n a_{ik}b_{kj}$$ in the case of $i=1$, we have $$[AB](1,j) = \sum_{k=1}^n a_{1k}b_{kj} = \\ \begin{cases} -2\cdot\frac{1(n+1 - j)}{n+1} + 1 \cdot \frac{2(n + 1 - j)}{n+1} & j \geq 2\\ -2 \cdot \frac{1(n+1-j)}{n+1} + 1 \cdot \frac{j(n + 1 - 2)}{n+1} & j=1 \end{cases} = \delta_{ij}$$ We can repeat a similar computation for the cases $2 \leq i \leq n-1$ and $i = n$.
Alternatively: we can find the characteristic polynomial of $A$ using induction (Pavel's method). From there, we could find the inverse using the Cayley Hamilton theorem.