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

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Detailed Solution

S=011101110⇒det⁡S=0(0−1)−1(0−1)+1(1−0) =0+1+1 =2The cofactors of elements of A are A11=+(0−1)=−1,A12=−(0−1)=1,A13=+(1−0)=1A21=−1(0−1)=1,A22=+(0−1)=−1,A23=−(0−1)=1A31=+(1−0)=1,A32=−(0−1)=1,A33=+(0−1)=−1Adj⁡A=A11A21A31A12A22A32A13A23A33=−1111−1111−1∴A−1=AdjAdet⁡A=12−1111−1111−1SAS−1=01110111012b+cc−ab−ac−bc+aa−bb−ca−ca+b12−1111−1111−1=140+c−b+b−c0+c+a+a−c0+a−b+a+bb+c+0+b−ac−a+0+a−cb−a+0+a+bb+c+c−b+0c−a+c+a+0b−a+a−b+0−1111−1111−1=1402a2a2b02b2c2c0−1111−1111−1=140+2a+2a0−2a+2a0+2a−2a−2b+0+2b2b+0+2b2b+0−2b−2c+2c+02c−2c+02c+2c+0=144a0004b0004c=a 0 00 b 00 0 c

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