91Ó°ÊÓ

Use the MATLAB code to compute \[A_4^{ - 1},A_5^{ - 1},A_6^{ - 1}\] as shown below:

\( > > {\mathop{\rm A}\nolimits} 4 = {\mathop{\rm ones}\nolimits} \left( 4 \right) - {\mathop{\rm eye}\nolimits} \left( 4 \right)\)

\({A_4} = \left[ {\begin{aligned}{*{20}{c}}0&1&1&1\\1&0&1&1\\1&1&0&1\\1&1&1&0\end{aligned}} \right]\)

\( > > {\mathop{\rm A}\nolimits} 5 = {\mathop{\rm ones}\nolimits} \left( 5 \right) - {\mathop{\rm eye}\nolimits} \left( 5 \right)\)

\({A_5} = \left[ {\begin{aligned}{*{20}{c}}0&1&1&1&1\\1&0&1&1&1\\1&1&0&1&1\\1&1&1&0&1\\1&1&1&1&0\end{aligned}} \right]\)

\( > > {\mathop{\rm A}\nolimits} 6 = {\mathop{\rm ones}\nolimits} \left( 6 \right) - {\mathop{\rm eye}\nolimits} \left( 6 \right)\)

\({A_6} = \left[ {\begin{aligned}{*{20}{c}}0&1&1&1&1&1\\1&0&1&1&1&1\\1&1&0&1&1&1\\1&1&1&0&1&1\\1&1&1&1&0&1\\1&1&1&1&1&0\end{aligned}} \right]\)

Use the MATLAB code to compute the inverse of \[{A_4},{A_5},{A_6}\] as shown below:

\( > > {\mathop{\rm inv}\nolimits} \left( {{\mathop{\rm A}\nolimits} 4} \right)\)

\(A_4^{ - 1} = \left[ {\begin{aligned}{*{20}{c}}{\frac{{ - 2}}{3}}&{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{{ - 2}}{3}}&{\frac{1}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{{ - 2}}{3}}&{\frac{1}{3}}\\{\frac{1}{3}}&{\frac{1}{3}}&{\frac{1}{3}}&{\frac{{ - 2}}{3}}\end{aligned}} \right]\)

\( > > {\mathop{\rm inv}\nolimits} \left( {{\mathop{\rm A}\nolimits} 5} \right)\)

\(A_5^{ - 1} = \left[ {\begin{aligned}{*{20}{c}}{\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{1}{4}}&{\frac{{ - 3}}{4}}\end{aligned}} \right]\)

\( > > {\mathop{\rm inv}\nolimits} \left( {{\mathop{\rm A}\nolimits} 6} \right)\)

\[A_6^{ - 1} = \left[ {\begin{aligned}{*{20}{c}}{\frac{{ - 4}}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}\\{\frac{1}{5}}&{\frac{{ - 4}}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}\\{\frac{1}{5}}&{\frac{1}{5}}&{\frac{{ - 4}}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}\\{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{{ - 4}}{5}}&{\frac{1}{5}}&{\frac{1}{5}}\\{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{{ - 4}}{5}}&{\frac{1}{5}}\\{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{1}{5}}&{\frac{{ - 4}}{5}}\end{aligned}} \right]\]

According to the construction of \({A_6}\) and the appearance of its inverse, the inverse is related to \({I_6}\). Moreover, \(A_6^{ - 1} + {I_6}\) is \(\frac{1}{5}\) times the \(6 \times 6\) matrix of ones.

Suppose \(J\) represents the \(n \times n\) matrix of ones. Then, the conjecture is \({A_n} = J - {I_n}\) and \(A_n^{ - 1} = \frac{1}{{n - 1}} \cdot J - {I_n}\).

Thus, the conjecture about the general form of \(A_n^{ - 1}\) is \({A_n} = J - {I_n}\) and \(A_n^{ - 1} = \frac{1}{{n - 1}} \cdot J - {I_n}\).