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Let x in R and let [{:(1,,1,,1),(0,,2,,2),(0,,0,,3):}],Q=[{:(2,,x,,x),(0,,4,,0),(x,,x,,6):}]and R=PQR^(-1) which of the following options is/are correct? |
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Answer» There exists a real, number x such that `PQ= QP` `P=[(1,1,1),(0,2,2),(0,0,3)],Q=[(2,x,x),(0,4,0),(x,x,6)]` `therefore""p^(-1)=(adj(P))/(|P|)` `as |P|=6 and " adj P"=[(6,0,0),(-3,3,0),(0,-2,2)]` `RARR""p^(-1)=(1)/(6)[(6,-3,0),(0,3,-2),(0,0,2)]` `therefore""|R|=|PQP^(-1)|""[because R=PQP^(-1)"(given)"]` `rArr""|R|=|P||Q||P^(-1)|=|Q|""[because|P||P^(-1)|=|I|=1]` `=|(2,x,x),(0,4,0),(x,x,6)|=|(2,x,x),(0,4,0),(x,x,5)|+|(2,x,0),(0,4,0),(x,x,1)|` `=|(2,x,x),(0,4,0),(x,x,5)|+2(4-0)-x(0-0)+0(0-4x)` `=|(2,x,x),(0,4,0),(x,x,5)|+8 " for all " x inR` `because""PQ=[(1,1,1),(0,2,2),(0,0,3)][(2,x,x),(0,4,0),(x,x,6)]` `=[(2+x,4+2x,x+6),(2x,2x+8,12),(3x,3x,18)]` `"and"QP=[(2,x,x),(0,4,0),(x,x,6)][(1,1,1),(0,2,2),(0,0,3)]` `=[(2,2+2x,2+5x),(0,8,8),(x,3,3x+18)]` There is no common value of 'x' for which each corresponding element of matrices PQ and QP is equal. For `x=0,Q=[(2,0,0),(0,4,0),(0,0,6)]` then, if `R[(1),(a),(b)]=6[(1),(a),(b)]` `rArrPQP^(-1)[(1),(a),(b)]=6[(1),(a),(b)]""[because R=PQP^(-1)]` `rArr(1)/(6)[(1,1,1),(0,2,2),(0,0,3)][(2,0,0),(0,4,0),(0,0,6)][(6,-3,0),(0,3,-2),(0,0,2)][(1),(a),(b)]=6[(1),(a),(b)]` `rArr""(1)/(6)[(2,4,6),(0,8,12),(0,0,18)][(6,-3,0),(0,3,-2),(0,0,2)][(1),(a),(b)]=6[(1),(a),(b)]` `rArr""[(12,6,4),(0,24,8),(0,0,36)][(1),(a),(b)]=36[(1),(a),(b)]` `rArr""[(12+6a+4b),(0+24a+8b),(0+0+36b)]=[(36),(36a),(36b)]` `rArr""6a+4b=24 and 12a=8b` `rArr""3a+2b=12 and 3a=2b` `rArr a=2 and b=3` So `a+b=5`. Now, `R[(alpha),(beta),(gamma)]=[(0),(0),(0)] and alphahati+betahatj+gammahatk" is a unit vector, so det "(R)=0` `rArrdet(Q)=0""[becauseR=PQP^(-1)" So, "|R|=|Q|]` `rArr""|(2,x,x),(0,4,0),(x,x,6)|=0` `rArr""2(24-0)-x(0-0)+x(0-4x)=` `rArr""48-4x^(2)=0` `rArr""x^(2)=12 rArr x= pm2sqrt3` So, for `x=1`, there does not exist a unit vector `alpha hat (i) +betahat(j) +gammahat(k)`, for which `R [{:(alpha,),(beta,),(gamma,):}]=[{:(0,),(0,),(0,):}]` Hence, options (b) and (d) are correct. |
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