Question

If A = $\left[\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right]$ and B = $\frac{1}{2} \left[\begin{array}{ccc}b+c&c-a&b-a\\c-b&c+a&a-b\\b-c&a-c&a+b\end{array}\right]$ then show that ABA⁻¹ is a diagonal matrix.

Answer 1

$A=\left[\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right] \\\\B=\frac{1}{2} \left[\begin{array}{ccc}b+c&c-a&b-a\\c-b&c+a&a-b\\b-c&a-c&a+b\end{array}\right]\\ \\AB=\left[\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right]\frac{1}{2} \left[\begin{array}{ccc}b+c&c-a&b-a\\c-b&c+a&a-b\\b-c&a-c&a+b\end{array}\right]\\\\[tex]AB=\frac{1}{2} \left[\begin{array}{ccc}0+c-b+b-c&0+c+a+a-c&0+a-b+a+b\\b+c+0+b-c&c-a+0+a-c&b-a+0+a+b\\b+c+c-b+0&c-a+c+a+0&b-a+a-b+0\end{array}\right]\\\\AB=\frac{1}{2}\left[\begin{array}{ccc}0&2a&2a\\2b&0&2b\\2c&2c&0\end{array}\right]\\\\AB=\left[\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0\end{array}\right] =A$

$So\\AB=A\\ABA^{-1}=AA^{-1}=I$

Which is a diagonal matrix.

But in this case, $A^{-1}$ does not exist because |A|=0 , and A is Singular.