InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Let `A=[a_("ij")]_(3xx3)` be a matrix such that `A A^(T)=4I` and `a_("ij")+2c_("ij")=0`, where `C_("ij")` is the cofactor of `a_("ij")` and `I` is the unit matrix of order 3. `|(a_(11)+4,a_(12),a_(13)),(a_(21),a_(22)+4,a_(23)),(a_(31),a_(32),a_(33)+4)|+5 lambda|(a_(11)+1,a_(12),a_(13)),(a_(21),a_(22)+1,a_(23)),(a_(31),a_(32),a_(33)+1)|=0` then the value of `lambda` is |
|
Answer» Correct Answer - 0.4 Given that `A A^(T)=4I` `implies |A|^(2)=4` or `|A|= pm 2` So `A^(T)=4A^(-1)=4 ("adj A")/(|A|)` `implies [(a_(11),a_(21),a_(31)),(a_(12),a_(22),a_(32)),(a_(13),a_(23),a_(33))]=4/(|A|)[(c_(11),c_(21),c_(31)),(c_(12),c_(22),c_(32)),(c_(13),c_(23),c_(33))]` Now `a_("ij")=4/(|A|) c_("ij")` `implies -2c_("ij")=4/(|A|) c_("ij")" "("as "a_("ij")+2c_("ij")=0)` `implies |A|=-2` Now `|A+4I|=|A+A A^(T)|` `=|A||I+A^(T)|` `=-2|(I+A)^(T)|` `=-2|I+A|` `implies |A+4I|+2|A+I|=0`, so on comparing, we get `5 lambda=2 implies lambda=2/5` |
|