If S=0 1 11 0 11 1 0,A=12b+cc−ab−ac−bc+aa−bb−ca−ca+b,then SAS–1=

S−1=12−1111−1111−1 Find SAS-10 1 11 0 11 1 012b+cc−ab−ac−bc+aa−bb−ca−ca+b12−1111−11111=a000b000c

If S=0 1 11 0 11 1 0,A=12b+cc−ab−ac−bc+aa−bb−ca−ca+b,then SAS–1=

S−1=12−1111−1111−1 Find SAS-10 1 11 0 11 1 012b+cc−ab−ac−bc+aa−bb−ca−ca+b12−1111−11111=a000b000c

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

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

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

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

$3 [\begin{array}{lll} a 0 0 \\ 0 b 0 \\ 0 0 c \end{array}]$

answer is A.

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Detailed Solution

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