InterviewSolution
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InterviewSolution
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यदि `A=[{:(,1,-2,1),(,-2,3,1),(,1,1,5):}]` हो, तो सत्यापित कीजिएः `-(adj A)^(-1)=adj A^(-1)`. |
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Answer» यहाँ `A=[{:(,1,-2,1),(,-2,3,1),(,1,1,5):}]` `therefore |A|=|{:(,1,-2,1),(,-2,3,1),(,1,1,5):}|` `=1(15-1)+2(-10-1)+1(-2-3)` `Rightarrow |A|=14-22-5=13 ne 0` `Rightarrow A^(-1)` का अस्तित्व है माना A में `a_(ij) `का सहखण्ड `A_(ij) `है, तब `A_(11)=(-1)^(1+1)|{:(,3,1),(,1,5):}|=15-1=14` `A_(12)=(-1)^(1+2)|{:(,-2,1),(,1,5):}|=-(-10-1)=11` `A_(13)=(-1)^(1+3)|{:(,-2,3),(,1,1):}|=-(-2-3)=5` `A_(21)=(-1)^(2+1)|{:(,-2,1),(,1,5):}|=-(-10-1)=11` `A_(22)=(-1)^(2+2)|{:(,1,1),(,1,5):}|=-(5-1)=4` `A_(23)=(-1)^(2+3)|{:(,1,-2),(,1,1):}|=-(1+2)=-3` `A_(31)=(-1)^(3+1)|{:(,-2,1),(,3,1):}|=-(2-3)=-5` `A_(32)=(-1)^(3+2)|{:(,1,1),(,-2,1):}|=-(1+2)=-3` `A_(33)=(-1)^(3+3)|{:(,1,-2),(,-2,3):}|=-(3-4)=-1` `therefore "adj A"=[{:(,14,11,-5),(,11,4,-3),(,-5,-3,-1):}]=[{:(,14,11,-5),(,11,4,-3),(,-5,-3,-1):}]` अंत: `A^(-1) =("adj A")/(|A|)=(1)/(-13)[{:(,14,11,-5),(,11,4,-3),(,-5,-3,-1):}]` `Rightarrow A^(-1) =[{:(,14/13,(-11)/13,5/13),(,(-11)/(13),(-4)/(13),(-3)/13),(,5/13,3/13,1/13):}]` `A^(-1)` के अवयवों का सहखण्ड है `C_11=(-1)^(1+1)|{:(,(-4)/(13),(3)/(13))/(,(3)/(13),(1)/(13)):}|=-(1)/(13)` `C_12=(-1)^(1+2)|{:(,(-11)/(13),(3)/(13))/(,(5)/(13),(1)/(13)):}|=(2)/(13)` `C_13=(-1)^(1+3)|{:(,(-11)/(13),(-4)/(13))/(,(5)/(13),(1)/(13)):}|=-(1)/(13)` `C_21=(-1)^(2+1)|{:(,(-11)/(13),(5)/(13))/(,(3)/(13),(1)/(13)):}|=(2)/(13)` `C_22=(-1)^(2+2)|{:(,(-14)/(13),(5)/(13))/(,(5)/(13),(1)/(13)):}|=-(3)/(13)` `C_23=(-1)^(2+3)|{:(,(-14)/(13),(-11)/(13))/(,(5)/(13),(3)/(13)):}|=-(1)/(13)` `C_31=(-1)^(3+1)|{:(,(-11)/(13),(5)/(13))/(,(-4)/(13),(3)/(13)):}|=-(1)/(13)` `C_32=(-1)^(3+2)|{:(,(-14)/(13),(5)/(13))/(,(-11)/(13),(3)/(13)):}|=-(1)/(13)` `C_33=(-1)^(3+3)|{:(,(-14)/(13),(-11)/(13))/(,(-11)/(13),(4)/(13)):}|=-(5)/(13)` `therefore "adj "A^(-1)=[{:(,-(1)/(13),(2)/(13),(-1)/(13)),(,(2)/(13),(-3)/(13),(-1)/(13)),(,(-1)/(13),(-1)/(13),(-5)/(13)):}]` `therefore "adj "A^(-1)=[{:(,-(1)/(13),(2)/(13),(-1)/(13)),(,(2)/(13),(-3)/(13),(-1)/(13)),(,(-1)/(13),(-1)/(13),(-5)/(13)):}]` अब: (adj A) `("adj A"^(-1))` `=[{:(,14,11,-5),(,11,4,-3),(,-5,-3,-1):}] [{:(,-(1)/(13),(2)/(13),(-1)/(13)),(,(2)/(13),(-3)/(13),(-1)/(13)),(,(-1)/(13),(-1)/(13),(-5)/(13)):}]` `Rightarrow ("adj A")("adj A"^(-1))=|{:(,1,0,0),(,0,1,0),(,0,0,1):}|=I_(3)` इसी प्रकार `("adj A"^(-1))("adj A)=I_(3)` `therefore (adj A")("adj A"^(-1))=I_(3)=(adj A^(-1)(adj A)` `Rightarrow ("adj A"^(-1))=("adj A"^(-1))=|{:(,-1/13,2/13,(-1)/13),(,2/13,(-3)/13,(-1)/13),(,(-1)/13,(-1)/13,(-5)/13):}|` यही सिद्ध करना था |
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