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Let A=|1sintheta1-sintheta1sintheta-1-si...

Let `A=|1sintheta1-sintheta1sintheta-1-sintheta1|,` where `0lt=thetalt=2pidot` Then `D e t(A)=0` (b) `D e t(A) in (2,oo)` `D e t(A) in (2,4)` (d) `D e t(A) in [2,4]`

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The correct Answer is:
(d)

`:." "A=[{:(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1):}]`
`rArr" "|A|=[{:(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1):}]`
`=1(1+sin^(2)theta)-sintheta(-sintheta+sintheta)+1(sin^(2)theta+1)=2+2sin^(2)theta`
`"For" 0 lethetale2pi`
`-1lesin thetale1 rArr0lesin^(2)thetale1`
`rArr" "1le1+sin^(2)thetale2`
`rArr" "2le2+sin^(2)thetale4rArr2le|A|le4`
`therefore" "detAin[2,4]`
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NAGEEN PRAKASHAN-DETERMINANTS-Miscellaneous Exercise
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