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Let |A|=|a(ij)|(3xx3) ne0 Each element ...

Let `|A|=|a_(ij)|_(3xx3) ne0` Each element `a_(ij)` is multiplied by by `k^(i-j)` Let `|B|` the resulting determinant, where `k_1 |A|+k_2 |B| =a` then `k_1+k_2 =`

A

`1`

B

`-1`

C

`0`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:
C

`(c )` `|A|=|{:(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33)):}|`
`|B|=|{:(a_(11),k^(-1)a_(12),k^(-2)a_(13)),(ka_(21),a_(22),k^(-1)a_(23)),(k^(2)a_(31),ka_(32),a_(33)):}|`
`=(1)/(k^(3))|{:(k^(2)a_(11),ka_(12),a_(13)),(k^(2)a_(21),ka_(22),a_(23)),(k^(2)a_(31),ka_(32),a_(33)):}|`
`=|A|`
`k_(1)|A|+k_(2)|B|=0`
`:.k_(1)+k_(2)=0`
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