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Bhabesh M.
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Linear Algebra
TutorMe
Question:

The 3x3 matrix A = [ 1 0 1], [0 2 -2], [0 0 3]. Find the value of A^29.

Bhabesh M.

The complexity of calculation and the power makes is too difficult to find the answer by manually multiplying matrices . But , Using the following result from diagonalization of matrices we have For any matrix A having distinct eigen values D= P^-1 A P. Where D is a diagonal matrix which has diagonal entries as the Eigen values of A and P is an invertible matrix which has Eigen vectors of A as Column vectors. Here our matrix has Eigen Values as 1,2 and 3 found out my simple calculation using the Characteristic polynomial as -λ^3+6*λ^2-11*λ+6 =0 so D=[1 0 0], [0 2 0], [0 0 3] and as P is Invertible so D= P^-1 A P = > P D P^-1 = A = > A = P D P^-1 = > A^29 = (P D P^-1 ) ^29 = > A^29 = P D^29 P^-1 { As (P D P^-1 ) ^29 = P D P^-1 * P D P^-1 *P D P^-1.............*P D P^-1} and P* P^-1 = I (identity matrix) So, A^29= P{ [1 0 0] , [0 2^29 0], [0 0 3^29] } P^-1. where P=[1 0 0], [0 1 0 ], [1/2 -2 1]

Set Theory
TutorMe
Question:

In a group of 60 people, 27 like Mathematics and 42 like Physics and each person likes at least one of the Subjects. How many like both Maths and Physics?

Bhabesh M.

Formula to use (Theory behind the problem): n( A U B ) = n(A) + n(B) - n(A and B) or n( A and B)= n(A) + n(B) - n( A U B) Solution : A= No of people who like Maths B= No of people who like Physics n( A and B)= n(A) + n(B) - n( A U B) n( A and B)= 27+ 42 - 60 n( A and B)= 9 So the number of people who like both maths and physics are 9.

Algebra
TutorMe
Question:

Does the ideal <x^2 +3> form a field as a quotient field over Q[x] (rationals)?

Bhabesh M.

An ideal forms a field as a quotient field over a Ring when it is the Maximal ideal. Here the ideal is generated by the polynomial x^2+3. The ideal here will be a maximal ideal iff the polynomial generating it is irreducible. So, the question boils down to decide whether the polynomial is irreducible over Q or not. Now, lets use Eisenstein's Criterion which states : Suppose we have the following polynomial with integer coefficients. P(x)=a{n}x^{n}+a{n-1}x^{n-1}+......... +a{1}x+a{0} If there exists a prime number p such that the following three conditions all apply: p divides each a{i} for i ≠ n, p does not divide a{n}, and p^2 does not divide a{0}, then P is irreducible over the rational numbers. choosing p=3 does the trick. Since, 3 | 3 =a{0} and p^2=9 doesn't divide 3=a{0} and 3 doesn't divide 1 = a{n}. So , Since the polynomial is irreducible over Q, it forms a maximal ideal over it and the quotient ring generated becomes a field

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