If S=0 1 11 0 11 1 0 and A=12b+c c−a b−ac−b c+a a−bb−c a−c a+b, then SAS−1=

∴S−1=AdjS|S|=12−1111−1111−1 Now SA =120 1 11 0 11 1 0b+c c+a b−cc−b c+a a−bb−c a−c a+b=0aab0bcc0SAS−1=0aab0bcc0⋅12−1111−1111−1=120 a ab 0 b0 0 c−1111−1111−1=122a0002b0002c=a 0 00 b 00 0 c

If S=0 1 11 0 11 1 0 and A=12b+c c−a b−ac−b c+a a−bb−c a−c a+b, then SAS−1=

∴S−1=AdjS|S|=12−1111−1111−1 Now SA =120 1 11 0 11 1 0b+c c+a b−cc−b c+a a−bb−c a−c a+b=0aab0bcc0SAS−1=0aab0bcc0⋅12−1111−1111−1=120 a ab 0 b0 0 c−1111−1111−1=122a0002b0002c=a 0 00 b 00 0 c

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If $S = [\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 0 \end{array}]$ and $A = \frac{1}{2} [\begin{array}{lll} b + c c - a b - a \\ c - b c + a a - b \\ b - c a - c a + b \end{array}]$ , then ${SAS}^{- 1} =$

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$[\begin{array}{lll} a 0 0 \\ 0 b 0 \\ 0 0 c \end{array}]$

$\frac{1}{2} [\begin{array}{lll} a 0 0 \\ 0 b 0 \\ 0 0 c \end{array}]$

$2 [\begin{array}{lll} a 0 0 \\ 0 b 0 \\ 0 0 c \end{array}]$

$[\begin{array}{lll} a b c \\ b c a \\ c a b \end{array}]$

answer is A.

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Detailed Solution

$∴ S^{- 1} = \frac{AdjS}{| S |} = \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]$

$Now SA = \frac{1}{2} [\begin{array}{lll} 0 1 1 \\ 1 0 1 \\ 1 1 0 \end{array}] [\begin{array}{lll} b + c c + a b - c \\ c - b c + a a - b \\ b - c a - c a + b \end{array}]$

$= [\begin{array}{ccc} 0 & a & a \\ b & 0 & b \\ c & c & 0 \end{array}]$

${SAS}^{- 1} = [\begin{array}{ccc} 0 & a & a \\ b & 0 & b \\ c & c & 0 \end{array}] \cdot \frac{1}{2} [\begin{array}{ccc} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{array}]$