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If A = [[a,b,c],[x,y,z],[p,q,r]], B= [[q...

If `A = [[a,b,c],[x,y,z],[p,q,r]], B= [[q , -b,y],[-p,a,-x],[r,-c,z]]` and if A is
invertible, then which of the following is not true?

A

`abs(A) = abs(B)`

B

`abs(A) = -abs(B)`

C

`abs(adjA) = abs(adjB)`

D

A is invertible `hArr` B is invertble

Text Solution

Verified by Experts

The correct Answer is:
A
`because abs(B) = abs((q, -b, y ),(-p, a ,-x),(r, -c, z))`
Applying `R-(2) rarr (-1) R_(2)`, then
`abs(B) = abs((q, -b, y ),(p, -a ,x),(r, -c, z))`
Appluing `C_(2) rarr (-1) C_(2),` then
`abs(B) = abs((q, b, y ),(p, a ,x),(r, c, z)) = abs(B^(T)) = abs((q,p,r),(b,a,c),(y,x,z))`
`= -abs((b,a,c),(q,p,r),(y,x,z)) [ R_(1) harr R_(2)]`
`= abs((b,a,c),(y,x,z),(q,p,r)) [ R_(1) harr R_(3)0]`
`= -abs((a,b,c),(x,y,z),(p,q,r)) =-abs(A)`
`rArr abs(B) = -abs(A)`
Also, `abs(adj B) = abs(B)^(2)`
` = abs(A)^(2) = abs(adjA) [because abs(A) ne 0 , "then"abs(B) ne 0]`
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