If S=\left[\begin{array}{ccc}0 & 1 & 1 \ 1 & 0 & 1 \ 1 & 1 & 0]\end{array} and A=\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}, then S A S^{-1}=

Correct Answer is : [a000b000c]

\left[\begin{array}{c}{\a\end{array} , 0 , 0 ; 0 , {\b} , 0 ; 0 , 0 , {\c}]}

\frac{1}{2}\left[\begin{array}{ccc}a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c]\end{array}

2\left[\begin{array}{ccc}a & 0 & 0 \ 0 & b & 0 \ 0 & 0 & c]\end{array}

\left[\begin{array}{ccc}a & b & c \ b & c & a \ c & a & b]\end{array}

AP EAPCET 15th May 2023 Shift 1 Solved Paper

Section: Mathematics

Question : 4 of 160

Marks: +1, -0

If S=[

] and A=‌

[

], then SAS−1=

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