Repeat the strategy of Exercise 33 to guess the inverse of \(A = \left( {\begin{aligned}{*{20}{c}}1&0&0& \cdots &0\\1&2&0&{}&0\\1&2&3&{}&0\\ \vdots &{}&{}& \ddots & \vdots \\1&2&3& \cdots &n\end{aligned}} \right)\). Prove that your guess is correct.

Suppose the inverse matrix of \(A\) is \(B\). Matrix \(B\) of order \(n \times n\) can be expressed as

\(B = \left( {\begin{aligned}{*{20}{c}}1&0&0& \cdots &0\\{ - \frac{1}{2}}&{\frac{1}{2}}&0&{}&{}\\0&{ - \frac{1}{3}}&{\frac{1}{3}}&{}&{}\\ \vdots &{}& \ddots & \ddots & \vdots \\0&0& \cdots &{ - \frac{1}{n}}&{\frac{1}{n}}\end{aligned}} \right)\).

For \(j = 1,...,n\), let \({{\bf{a}}_j}\), \({{\bf{b}}_j}\) and \({{\bf{e}}_j}\) denote the \(j\)th columns of \(A\), \(B\) and \(I\), respectively.

For \(j = 1,...,n - 1\),

\({{\bf{a}}_j} = j\left( {{{\bf{e}}_j} + ....... + {{\bf{e}}_n}} \right)\),

\({{\bf{b}}_j} = \frac{1}{j}{{\bf{e}}_j} - \frac{1}{{j + 1}}{{\bf{e}}_{j + 1}}\), and

\({{\bf{b}}_n} = \frac{1}{n}{{\bf{e}}_n}\).

\(\begin{aligned}{c}A{b_j} = A\left( {\frac{1}{j}{{\bf{e}}_j} - \frac{1}{{j + 1}}{{\bf{e}}_{j + 1}}} \right)\\ = \frac{1}{j}{{\bf{a}}_j} - \frac{1}{{j + 1}}{{\bf{a}}_{j + 1}}\\ = \left( {{{\bf{e}}_j} + ..... + {{\bf{e}}_n}} \right) - \left( {{{\bf{e}}_{j + 1}} + .... + {{\bf{e}}_n}} \right)\\ = {{\bf{e}}_j}\end{aligned}\)

So, \(AB = I\).

As \(\frac{1}{n}{a_n} = {e_n}\),

\(\begin{aligned}{c}B{{\bf{a}}_j} = j\left( {B{{\bf{e}}_j} + ..... + B{{\bf{e}}_n}} \right)\\ = j\left( {{{\bf{b}}_j} + ..... + {{\bf{b}}_n}} \right)\\ = j\left( {\frac{1}{j}{{\bf{e}}_j}} \right)\\ = {{\bf{e}}_j}.\end{aligned}\)

So, \(BA = I\).

The inverse of matrix \(A\) is \(\left( {\begin{aligned}{*{20}{c}}1&0&0& \cdots &0\\{ - \frac{1}{2}}&{\frac{1}{2}}&0&{}&{}\\0&{ - \frac{1}{3}}&{\frac{1}{3}}&{}&{}\\ \vdots &{}& \ddots & \ddots & \vdots \\0&0& \cdots &{ - \frac{1}{n}}&{\frac{1}{n}}\end{aligned}} \right)\) for which \(AB = BA = I\).