Q.1. If θ  ∅ = an odd multiple of — find the product of matrices of find the product of matrices
Ans.The product of A and B is
Since d and S differ by an odd multiple of therefore θ  ∅ = an odd multiple of
thus, cos (00) = 0. Hence,
Q.2. Find the values of a, b, c, d, e, f where a, b, c and f are positive and d and e are negative such that the matrix is orthogonal.
Ans. In an orthogonal matrix the sum of the squares of elements in any row or any columns is unity. Here,
⇒ b^{2} + c^{2} = 5 and b+2c = 4
solving these two relation we obtain b = 2 and c = 1
Q.3. If A is a matrix such that
Find the matrix A^{20} + A^{15} + 2I, where I is 3 x 3 Identity matrix.
Ans.
Thus, A^{2} = I, A^{3} = A, A^{4} = I, A^{5} = A and so on.
Thus we see that the odd power of A are equal to A and even powers of A are equal to I. Hence A^{20} + A^{15} + 21 = I + A + 2I = A + 3I
Q.4. The trace of a 3 x 3 matrix is 12 and the determinant of the matrix is 48 . The product of the smallest and the largest eigenvalues is 12. If all the eigenvalues of the matrix are different then
(a) Find the eigenvalues of the matrix.
(b) Write the chatacteristics equation.
Ans. λ_{1} + λ_{2} + λ_{3} = 12, λ_{1}λ_{2}λ_{3} = 48
Let A be the smallest eigenvalue and λ_{3} be the largest eigenvalue .
Then, λ_{1}λ_{3} = 12 ⇒ λ_{2 }=
Hence, λ_{1 }+ λ_{3} = 8
(λ_{1}  λ_{3})^{2} = (λ_{1} + λ_{3})^{2}  4λ_{1}λ_{3} = 8^{2} 4.12 =16 ⇒ λ_{1}  λ_{3 }= + 4
λ_{1}= 2 and λ_{3} = 6
(b) The characterstic equation is
(λ  2)(λ  4)(λ  6) = 0 ⇒ λ^{3} 12λ^{2} + 44λ  48 = 0
Q.5. Find the points where the curve represented by [ x y ] cuts the x and y axes.
Ans.
⇒ x (5x  7 y) + y (7 x + 3 y ) = 30
5x^{2} + 3 y^{2} + ( 7 xy 7 xy) = 30
This is an equation of ellipse. Its graph is shown in the figure.
Q.6. Given the matrix B =
(a) Write the characteristic polynomial.
(b) If one of the eigenvalues of the matrix is twice repeated , find all the eigenvalues.
(c) Find the eigenvector associated with each eigenvalue.
Ans. (a) Tr (B) = 3  5 + 2 = 0, B = 16
Cofactors of diagonal elements are
B_{11} = 4, B_{22} = 0, B_{33} = 8
Therefore, P (A) = λ^{3} 12λ +16
(b) Since one of the eigenvalues of the matrix is twice repeated hence P (λ) and P'(λ) have the same roots.
P'(λ) = 3λ^{2} 12 ⇒ P'(λ) = 0 gives λ = ±2
Putting A = 2, makes P (λ) = 0 . Therefore λ = 2 is a twice repeated root.
From the property of roots we must have λ_{1}λ_{2}λ_{3} = 16 ⇒ 2 • 2 • λ = 16 ⇒ λ = 4. Thus λ = 2 is a twice repeated root and λ = 4 is another root.
(c) Subtracting λ = 2 down the main diagonal of B gives the matrix
Here we have subtracted row 3 from row 2.
x_{1}  x_{2} + x_{3} = 0 and x_{3} = 0
This system has only one free variable hence is an eigenvector.
Q.7. For the matrix
Ans. We see that A^{2} = A, i.e, A is an idempotent matrix. Therefore
Q.8. Given the matrix
(a) Find the eigenvalue of matrix B .
(b) Find the eigenvectors of matrix B .
(c) Diagonalize matrix B .
(d) Find the matrix
(e) Using CayleyHamilton’s theorem find the characteristic equation satisfied by matrix B .
(f) Are the eigenvectors of matrix B linearly independent? Are the eigenvectors orthogonal to each other?
Ans. (a) eigenvalves of B
⇒ (1  λ)( 2  λ)(3  λ) = 0
⇒ λ = 1,λ = 2 and λ = 3
(b) Eigenvectors of B
x_{1} + x3 = x_{1} ⇒ x_{3} = 0
2x_{2} = x_{2} ⇒ x_{2} = 0
x_{2} + 3x_{3} = x_{3}
For λ = 2,
x_{1} + x_{3} = 2x_{1} ⇒ x_{3} = x_{1}
2x_{2} = 2x_{2}
x_{2} + 3x_{3} = 2x_{3} x_{2} = x_{3}
x_{1} + x_{3} = 3x_{1} ⇒ x_{3} = 2x_{1}
2x_{2} = 3x_{2} ⇒ x_{2} = 0
x_{2} + 3x_{3} = 3x_{3}
(c) We can take as three eigenvectors corresponding to λ = 1,λ = 2 and λ = 3. The matrix P diagonalizing matrix B is
Hence P^{1}AP = D, where D =
(d) we can write B = PDP^{1}
(e) The characteristic polynomial of the matrix is
P(λ) = (λ1)(λ2)(λ3) = λ^{3} 6λ^{2} + 11λ  6
Hence, according to CayleyHamiltoris theorem matrix equation satisfied by matrix B is B^{3}  6B^{2} + 11B  6I = 0
(f) We know that the eigenvectors belonging to distinct eigenvalves are linearly independent . Hence the eigenvalves of B corresponding to λ = 1,λ = 2 and λ = 3 are linearly independent. are the eigenvectors corresponding to λ = 1,λ = 2 and λ = 3 respectively.
From orthogonality condition
which is possible only when either k = 0 or a = 0. But k = 0 and a = 0 is not allowed. are not orthogonal. Similarly we see that no pair of eigenvectors are orthogonal.
Q.9. If A and B are two matrices such that AB and A + B are both defined. What can be said about the relation between the order of the two matrices? Are A and B square?
Ans. Let A be an m x n matrix. Since A + B is defined, therefore B should also be an m x n . Further since AB in defined, hence m = n , hence A and B are square matrices of the same order.
Q.10. Let A, B, C and D be four non zero matrices such that A =
If AB = CD , find matrix B.
Ans.
Since AB = CD ⇒ AB = I
Hence matrix B is the inverse of matrix A.
Q.11. If A is a matrix such that A = then
(a) Find matrix A^{100} + A^{99} + A^{98} + A^{97} + A^{2} + A .
(b) Find the Trace of matrix in part (a).
(c) Find the Determinant of matrix in part (a).
Ans. (a)
⇒ A^{3 }= AA^{2} = A^{2} = A, A^{4} = A^{2} = AA^{5} = A^{2} = A....
Thus we have see that all powers of A are equation to A, hence we can write
A^{100} + A^{99} + A^{98} + A^{97} + ••• A^{2} + A
= A + A + A + ••• upto hundred times
(b) The Trace of A^{100} + A^{99} + A^{98} + A^{97} + ••• A^{2} + A = 200+300  300 = 200
(c) The determinant of A^{100} + A^{99} + A^{98} + ••• A^{2} + A is
= 10^{6} [2 (9 [ 8) + 2 (3  4)4 (2  3)] = 0
Q.12. If A and B are two matrices such that AB = A and BA = B . If A^{2} + B^{2} = 2α (A + B) then find the value of α .
Ans. We have AB = A
⇒ A (BA) = A[∴B = BA]
AB)A = A = A^{2} = A (i)
Again BA = B ⇒ BAB = B (Since A = AB)
(BA)B = B ⇒ B^{2} = B (ii)
Adding equations (i) and (ii)
A^{2} + B^{2} = A + B (iii)
From the question
A^{2} + B^{2} = 2α (A + B) (iv)
From (iii) and (iv)
Q.13. If
(a) Find Tr (A) + Tr (B ) + Tr (C ).
(b) Find Tr (A^{T} ) Tr (B^{T }).Tr (C^{T} ).
(c) Find Tr (A^{1} ).Tr ( B^{1}) Tr (C^{1}).
(d) Find the product of determinant of three matrices.
Ans. A = (8  0)2 (0  0) + 3 (0  0) = 8
B = 2 (12  0) = 24
C =10
Since the eigenvalues of a triangular or diagonal matrix are just the diagonal elements hence
(a) Tr ( A) + Tr (B) + Tr (C ) = 7 + 9 + 8 = 24
(b) Tr (A^{T})  Tr (B^{T})• Tr (C^{T} ) = 7 • 9 • 8 = 504
(c) Tr (A^{1})• Tr (B^{1})• Tr (C^{1}) =
(d) det ( A)• det ( B )• det (C)
=8 x 24 x 10
= 1920
Q.14. Given the matrices A =
(a) Calculate B^{1 }AB and B^{1}A^{1}B.
(b) What is the relation between the two matrices of part (a)
Ans.
A = [0  2] +1[0 + 4] + [8  0] = 6
B = 1 + 0 + 1 x 0 = 1
(a)
And, D = B^{1} A^{1}B =
Q.15. Let .If A = 3 is one of the eigenvalues of the matrix then
(a) Find other eigenvalues of the matrix.
(b) To each of the eigenvalues of the matrix find the associated eigenvectors.
Ans. (a) The characteristic polynomial of the given matrix is
P(λ) = λ^{3} Tr(A)λ^{2} +(A_{11} + A_{22} + A_{33})λ 
Where Tr (A) is the trace of matrix A. A_{11}, A_{22} and A_{33} are cofactors of diagonal elements and is the determinant of matrix A .
Here, Tr(A) = 4 + 5 + 2 = 11 and A = 4(l2)1(6)1(3) = 45
⇒ P (2) = λ^{3}11λ^{2} + 39λ45
Since A = 3 is a root of this polynomial hence (A3) is a factor of P(λ). Dividing P(λ) by A3 we obtain
P(λ) = (λ3)(λ^{2} 8λ +15) = (λ3)(λ3)(λ5)
Thus, λ = 3 is a repeated eigenvalue. And λ = 3 is another eigenvalue.
(b) Subtracting λ = 3 down the diagonal of matrix A we obtain the matrix
Hence we can take (1,1,0) and (1,0,1) as two linearly independent eigenvalues. Subtracting λ = 5 down the matrix diagonal gives the matrix.
Thus, x_{1}  x_{3} = 0
X_{2}  2 x_{3} = 0
Here we have only one free variable, hence (1,2,1) is an eigenvector corresponding to A = 5.
Q.16. For the matrix
(a) Find the eigenvalues and eigenvectors .
(b) Find the matrix P which digonalizes matrix A .
(c) Write the diagonal form of A .
(d) Find the Trace and determinant of matrix e^{2A} .
Ans. (a) Given
Solving above matrix for eigenvalues:
⇒ (2  λ)(1  λ)(3  λ) = 0
λ = 1, λ_{2} = 3, λ_{3} = 2
Eigenvectors corresponding to respective eigenvalues, on solving we get as:
(b) Since we have three linearly independent eigenvectors, we can digonalise matrix A . Putting the eigenvectors of A as column vectors , we obtain
(c) Now, inverse of P is
Thus, Diagonalised matrix of given matrix A is
D = P^{1}AP
(d) The eigenvalues of matrix e^{2A} are e^{2},e^{4}, and e^{6}.
Hence the trace of e^{2A} is e^{2} (1 + e^{2} + e^{4})
The determinant of e^{2A} is e^{2}.e^{4}.e^{6} = e^{12}
Q.17. For the Pauli’s spin matrices
(a) Find the matrix
(b) Find the inverse of each of Pauli’s spin matrices.
(c) Find the eigenvalves of each Pauli’s spin matrix.
(d) Find the normalized Eigenvectors of each Pauli’s spin matrix.
(e) Diagonalize each of the Pauli’s matrices.
(f) Find the trace and determinant of matrix
(g) Find the matrices and
(h) Find the trace and determinant of matrix
Ans. (a)
(b) From the above relation we see that each Pauli’s spin matrix its
own inverse. A matrix which is its own inverse is called involutory. Thus each Pauli’s spin matrix is involutory.
(c) Eigenvalues of
Eigenvalues of
Eigenvalues of
We see that the Eigenvalues of each Pauli’s spin matrices are ±1.
(d) Eigenvectors of σ_{1}:
For λ = 1, we have ,
For λ = 1 ,we have,
Eigen vectors of σ_{2}
For λ = 1, we have ,
For λ = 1, we have ,
Eigen vectors of σ_{3} are,
(e) Diagonalisation of σ_{1},
Diagonalisation of σ_{2},
σ_{3} is already diagonalised.
(g)
Now, Since σ_{3} is diagonal we obtain
(h) σ_{1}σ_{2}σ_{3}
As, σ1σ2 = iσ3 ⇒ σ_{1}σ_{2}σ_{3} =
Trace λ_{1} + λ_{2} = i + i = 2i
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