InterviewSolution
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InterviewSolution
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आव्यूह `A=[{:(,cos alpha,-sin alpha,0),(,sin alpha,cos alpha,0),(,0,0,1):}]` के लिए सत्यापित कीजिएः -A(adj A)=|A| I. |
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Answer» यहाँ `A=[{:(,cos alpha,-sin alpha,0),(,sin alpha,cos alpha,0),(,0,0,1):}]` माना A में अवयव `a_(ij)` का सहखण्ड `A_(ij)` है तब `A_(11)-|{:(cos alpha,0),(0,1):}|=cos alpha` `A_(12)=-|{:(,sin alpha, 0),(,0,1):}|=-(sin alpha-0)=-sin alpha` `A_(13)=-|{:(,sin alpha, cos alpha),(,0,0):}|=0` `A_(21)=-|{:(,sin alpha,0),(,0,1):}|=-(sin alpha-0)=-sin alpha` `A_(22)=-|{:(,cos alpha,0),(,0,1):}|=cos alpha` `A_(23)=-|{:(,cos alpha,-sin alpha),(,0,1):}|=0` `A_(31)=-|{:(,-sin alpha,0),(,cos alpha,0):}|=0` `A_(32)=-|{:(,-cosalpha,0),(,sin alpha,0):}|=0` `A_(33)=-|{:(,cos alpha,-sin alpha),(,sin alpha, cos alpha):}|=cos^(2)alpha+sin^(2)alpha=1` `therefore adj A=[{:(,A_11,A_12,A_13),(,A_21,A_22,A_23),(,A_31,A_32,A_33):}]` `therefore adj A=[{:(,A_11,A_21,A_31),(,A_12,A_22,A_32),(,A_13,A_23,A_33):}]` `therefore adj A=[{:(,cos alpha,sinalpha,0),(,-sin alpha,cos alpha,0),(,0,0,1):}]` अब `|A|=|{:(,cos alpha,-sin alpha,0),(,sin alpha,cos alpha,0),(,0,0,1):}|` `=1xx |{:(,cos alpha,-sin alpha),(,sin alpha,cos alpha):}|=cos^(2)alpha+sin^(2)alpha=1` और `"A.(adj A)"=[{:(,cos alpha,-sin alpha,0),(,sin alpha,cos alpha,0),(,0,0,1):}][{:(,cos alpha,sin alpha,0),(,-sin alpha,cos alpha,0),(,0,0,1):}]` `Rightarrow "A.(adj A)"=[{:(,1,0,0),(,0,1,0),(,0,0,1):}]=I` `Rightarrow "A.(adj A)"=1.I=|A|I.` यही सिद्ध करना था |
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