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There are two possible values of A in the solution of the matrix equation `[[2A+1,-5],[-4,A]]^(-1) [[A-5,B],[2A-2,C]]= [[14,D],[E,F]]`, where A, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, isA. `8/3`B. `11/3`C. `1/3`D. `19/3` |
Answer» Correct Answer - D `because [[A-5,B],[2A-2,C]]= [[2A+1,-5],[-4,A]][[14,D],[E,F]]` `rArr A-5 = 28 A + 14 - 5E` `rArr 5e = 27 A + 19` …(i) `2A - 2 = -56 + AE` ` rArr AE = 2A +54 ` (ii) From eq. (i), we get `5AE = 27A^(2) + 19A` `rArr 5 (2A+54)=27 A^(2) + 19A ` [from Eq. (ii) ] `rArr 27A^(2) + 9 A - 270 = 0` `rArr 9 (A-3) (3A+10)=0` `therefore A= 3, A= -10/3` `therefore` Absolute value of difference `=abs(3+10/3 ) = 19/3` |
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