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# Tutor profile: Rohit Y.

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Rohit Y.
Tutor for 3+ years.
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## Questions

### Subject:Linear Algebra

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Question:

T:R^{4} ---> R^{4}, defined by T(e_{1})=e_{2}, T(e_{2})=e_{3}, T(e_{3})=0, T(e_{4})=e_{3}. Where e_(i)'s are members of the basis, Then (a) T is nilpotent (b) T has at least one non-zero eigen value (c) index of nilpotent is three (d) T is not nilpotent

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Rohit Y.

Answer is that: T is nilpotent with at least one nonzero eigen value and index of nilpotent is three! Hint: construct a matrix of T corresponding to the given linear transformation( if you take standard basis, to form a matrix, you will notice that all the eigen values are zero with certain power of matrix is zero. So clearly option (d) is not the case. For other option one needs to go with generalized proof, keeping in mind all the definitions required!

### Subject:Pre-Algebra

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Question:

The equation mx - 8 = 6 - 7(x + 3) DOES NOT have any solution if m = A. 3 B. 7 C. -7 D. 0

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Rohit Y.

Answer is -7! As if we simplify above equation (given in question) and collect variable x on one side, it will become: mx-7x = 7, clearly for the value of m = -7, x vanishes and we will not be able to find its value!

### Subject:Algebra

TutorMe
Question:

A group of order 505 is cyclic?

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Rohit Y.

Group of order 101 is unique. Since that subgroup is unique, it is normal. So G has a normal subgroup of order 101 and a subgroup of order 5 (which might not be normal). If you consider Sylow theorems, you will be able to know that the groups of this type are called semidirect products determined by maps from Z/5 into Aut(Z/101)≃Z/100. There is a map from Z/5 into Z/100, and so we have a noncyclic group of order 505. To be more precise, it is: ⟨x,y:x5=1,y101=1,xy=y36x⟩

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