InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
निम्नलिखित में दिए गए प्रत्येक आव्यूहों के व्युत्क्रम (जिनका अस्तित्व हो) ज्ञात कीजिए।`[(1,0,0),(0,cos alpha,sin alpha),(0, sin alpha,-cos alpha)]` |
|
Answer» माना `A=[(1,0,0),(0,cos alpha,sin alpha),(0,sin alpha,-cos alpha)]` `implies|A|=+|(1,0,0),(0,cos alpha,sin alpha),(0, sin alpha, -cos alpha)|` `=1(-cos^(2)alpha-sin^(2)alpha)-0+0=-1!=0` `A_(11)=(-1)^(2)(-cos^(2)alpha-sin^(2)alpha)=-1` `A_(12)=(-1)^(3)(0-0)=0` `A_(13)=(-1)^(4)(0-0)=0`, `A_(21)=(-1)^(3)(0-0)=0` `A_(22)=(-1)^(4)(-cos alpha-0)=-cos alpha` `A_(23)=(-1)^(5)(sin alpha-0)=-sin alpha` `A_(31)=(-1)^(4)(0-0)=0` `A_(32)=(-1)^(5)(sin alpha-0)=-sin alpha` `A_(33)-(-1)^(6)(cos alpha -0)=cos alpha` `:. adj A=[(-1,0,0),(0,-cos alpha, -sin alpha),(0,-sin alpha, cos alpha)]` `=[(-1,0,0),(0,-cos alpha, -sin alpha),(0,-sin alpha,cos alpha)]` अब `A^(-1)=1/(|A|)adjA=1/(-1)[(-1,0,0),(0,-cos alpha,-sin alpha),(0,-sin alpha, cos alpha)]` `=[(1,0,0),(0, cos alpha, sin alpha),(0,sin alpha, -cos alpha)]` |
|