Use the algorithm from this section to find the inverse of

\(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}\end{aligned}} \right)\)and \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{1}}&{\bf{1}}\end{aligned}} \right)\)

Let \(A\) be the corresponding \(n \times n\) matrix, let \(B\) be its inverse. Guess the form of \(B\) and then prove that \(AB = I\) and \(BA = I\).

Form the matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\1&1&0\\1&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\)

At row three, subtract row one from row three, i.e., \({R_3} \to {R_3} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\1&1&0\\0&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\{ - 1}&0&1\end{aligned}} \right)\)

At row two, subtract row one from row two, i.e., \({R_2} \to {R_2} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\{ - 1}&0&1\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\,\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\)

So, the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\1&1&1&0\\1&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{aligned}} \right)\)

At row four, subtract row one from row four, i.e., \({R_4} \to {R_4} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\1&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

At row three, subtract row one from row three, i.e., \({R_3} \to {R_3} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\1&1&0&0\\0&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\{ - 1}&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

At row two, subtract row one from row two, i.e., \({R_2} \to {R_2} - {R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&1&1&0\\0&1&1&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\{ - 1}&0&1&0\\{ - 1}&0&0&1\end{aligned}} \right)\)

At row four, subtract row three from row four, i.e., \({R_4} \to {R_4} - {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&1&1&0\\0&0&0&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\{ - 1}&0&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{aligned}\,\,\,\,\,\,\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\)

So, the inverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\).

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

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

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

\({{\bf{a}}_j} - {{\bf{a}}_{j + 1}} = {{\bf{e}}_j}\),

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

\({{\bf{a}}_n} = {{\bf{b}}_n} = {{\bf{e}}_n}\).

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

So, \(AB = I\).

As \({a_n} = {b_n} = {e_n}\),

\(\begin{aligned}{c}B{{\bf{a}}_j} = B\left( {{{\bf{e}}_j} + ..... + {{\bf{e}}_n}} \right)\\ = {{\bf{b}}_j} + ..... + {{\bf{b}}_n}\\ = \left( {{{\bf{e}}_j} - {{\bf{e}}_{j + 1}}} \right) + \left( {{{\bf{e}}_{j + 1}} - {{\bf{e}}_{j + 2}}} \right) + ..... + \left( {{{\bf{e}}_{n - 1}} - {{\bf{e}}_n}} \right) + {{\bf{e}}_n}\\ = {{\bf{e}}_j}.\end{aligned}\)

So, \(BA = I\).

The inverse of the given matrices are \(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 1}&1&0\\0&{ - 1}&1\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}1&0&0&0\\{ - 1}&1&0&0\\0&{ - 1}&1&0\\0&0&{ - 1}&1\end{aligned}} \right)\), and \(AB = BA = I\).