If $S=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]$ and…
If $S=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]$ and $A=\frac{1}{2}\left[\begin{array}{lll}b+c & c-a & b-a \\ c-b & c+a & a-b \\ b-c & a-c & a+b\end{array}\right]$, then $\mathrm{SAS}^{-1}=$
- $\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$
- $\frac{1}{2}\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$
- $2\left[\begin{array}{lll}\mathrm{a} & 0 & 0 \\ 0 & \mathrm{~b} & 0 \\ 0 & 0 & \mathrm{c}\end{array}\right]$
- $\left[\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right]$
Solution
$\mathrm{S}^{-1}=\frac{1}{2}\left[\begin{array}{ccc}-1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]$ (obtainded from matrix ' $\mathrm{S}$ ') Consider SA $=\frac{1}{2}\left[\begin{array}{ccc}0 & 2 \mathrm{a} & 2 \mathrm{a} \\ 2 \mathrm{~b} & 0 & 2 \mathrm{~b} \\ 2 \mathrm{c} & 2 \mathrm{c} & 0\end{array}\right]$
Hence SAS $^{-1}=(\mathrm{SA}) \mathrm{S}^{-1}$
$
=\frac{1}{4}\left[\begin{array}{ccc}
4 a & 0 & 0 \\
0 & 4 b & 0 \\
0 & 0 & 4 c
\end{array}\right]=\left[\begin{array}{lll}
a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c
\end{array}\right]
$
Asked in: AP EAMCET 2023 (15 May Shift 1)
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