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If A = [[a,b],[c,d]] (where bc ne0 ) sat...

If `A = [[a,b],[c,d]]` (where bc `ne0` ) satisfies the
equations `x^(2) + k = 0,` then

A

a + d = 0

B

`k=-abs(A)`

C

`k=abs(A)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A, C
We have , `A^(2) = [[a,b],[c,d]][[a,b],[c,d]]= [[a^(2) + bc , ab + bd],[ac+cd, bc+d^(2)]]`
As A satisfies `X^(2) + k = 0,` therefoer
`rArr [[a^(2) + bc +k, (a+d) b],[(a+d)c, bc+d^(2)+k]]= [[0,0],[0,0]]`
`rArr a^(2) + bc + k = 0, (a + d) b=0, `
`(a+b) c = 0 and bc + d^(2) + k = 0`
As `bcne 0 rArr b ne 0 , c ne 0 `
So, `a + d = 0 rArr a = -d`
Also `k = - (a^(2) + bc) = - (-ad+bc) = (ad + bv) = (ad - bc) = abs(A)`
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