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If `A=[(alpha,2),(2,alpha)]` and determinant `(A^3)=125,` then the value of `alpha` is (a) `+-1` (b) `+-2` (c) `+-3` (d) `+-5`A. `pm1`B. `pm2`C. `pm3`D. `pm5` |
Answer» Correct Answer - C We have, `absA={:abs((alpha,2),(2,alpha))=alpha^2-4:}` `:. Abs(A^3)=125` `rArr abs(A)^3=125` `rArr(alpha-4)^3-125rArr alpha^2=9rArr alpha=pm3`. |
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