Whether the given matrix is Hermitian or Skew-Hermitian or Unitary. The given matrix is Hermitian _ . Given: The given matrix is | 6 i − i 6 | . Definition used: Hermitian, Skew-Hermitian and Unitary: Hermitian: if A ¯ T = A that is, a ¯ k j = a j k . Skew-Hermitian: if A ¯ T = − A that is, a ¯ j k = − a j k . Unitary: if A ¯ T = A − 1 . Procedure used: The value of A ¯ = [ a ¯ j k ] is obtained from A= [ a j k ] by replacing each entry a j k = α + i β (α, β real) with its complex conjugate a ¯ j k = α − i β . Also, A ¯ T = [ a ¯ k j ] is the transpose of A ¯ , hence the conjugate transpose of A. Calculation: Consider the given matrix A= | 6 i − i 6 | . Obtain the value of A ¯ by using above procedure. A= | 6 i − i 6 | A ¯ = | 6 − i i 6 | Obtain the value of A ¯ T by using above procedure. A ¯ = | 6 − i i 6 | A ¯ T = | 6 i − i 6 | A ¯ T = A Thus, the given matrix is Hermitian _ . The eigen values of | 6 i − i 6 | . The eigen values are λ = 7 or λ = 5 _ . Given: The given matrix is | 6 i − i 6 | . Definition used: Let A be a n × n matrix. The eigen values of the matrix A is obtained by solving the characteristic equation D ( λ ) = 0 that is det ( A − λ I ) = 0 , where λ is the eigen value. Calculation: Consider the given matrix A= | 6 i − i 6 | . Obtain the eigenvalues of A as follows. It is known that the eigenvalues of A are the roots of the characteristic equation det ( A − λ I ) = 0 . det ( | 6 i − i 6 | − λ | 1 0 0 1 | ) = 0 det ( | 6 i − i 6 | − | λ 0 0 λ | ) = 0 det ( | 6 − λ i − i 6 − λ | ) = 0 ( 6 − λ ) ( 6 − λ ) − ( − i 2 ) = 0 On further simplification, λ 2 − 12 λ + 36 − 1 = 0 λ 2 − 12 λ + 35 = 0 ( λ − 7 ) ( λ − 5 ) = 0 λ = 7 or λ = 5 Thus, the eigen values are λ = 7 or λ = 5 _ . The eigen vector | 6 i − i 6 | . The eigen vector are is x = [ − i 1 ] , x = [ i 1 ] _ . Consider the given matrix A= | 6 i − i 6 | . Obtain the eigenvectors as follows. Substitute λ = 5 in [ A − λ I ] as, | 6 i − i 6 | − ( 5 ) | 1 0 0 1 | = | 6 i − i 6 | − | 5 0 0 5 | = | 6 − 5 i − i 6 − 5 | = | 1 i − i 1 | Reduce the matrix as follows. [ A − λ I ] = | 1 i − i 1 | = | 1 i 0 0 | ( R 2 → R 2 + i R 1 ) Let x = [ x 1 x 2 ] be the eigen vector. Solve the equation [ A − λ I ] x = 0 as follows: [ A − λ I ] x = 0 [ 1 i 0 0 ] [ x 1 x 2 ] = 0 x 1 = − i x 2 ( Let x 2 = 1 ) x 1 = − i Thus, the vector x can be written as follows: x = [ x 1 x 2 ] x = [ − i 1 ] Therefore, the eigen vector for λ = 5 is x = [ − i 1 ] _ . Substitute λ = 7 in [ A − λ I ] as, | 6 i − i 6 | − ( 7 ) | 1 0 0 1 | = | 6 i − i 6 | − | 7 0 0 7 | = | 6 − 7 i − i 6 − 7 | = | − 1 i − i − 1 | Reduce the matrix as follows. [ A − λ I ] = | − 1 i − i − 1 | = | − 1 i 0 0 | ( R 2 → R 2 − i R 1 ) Let x = [ x 1 x 2 ] be the eigen vector. Solve the equation [ A − λ I ] x = 0 as follows: [ A − λ I ] x = 0 [ − 1 i 0 0 ] [ x 1 x 2 ] = 0 − x 1 = − i x 2 ( Let x 2 = 1 ) x 1 = i Thus, the vector x can be written as follows: x = [ x 1 x 2 ] x = [ i 1 ] Therefore, the eigen vector for λ = 5 is x = [ i 1 ] _ .

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1. 2. Show that the following are not logically equivalent by finding a counterexample: (p^q) →r and (db) V (d←d) Show that the following is not a contradiction by finding a counterexample: (pV-q) AqA (pv¬q Vr) 3. Here is a purported proof that (pq) ^ (q → p) = F: (db) v (bd) = (db) v (bd) =(qVp) A (g→p) = (¬¬q V ¬p) ^ (q→ p) (db) V (db) = =¬(a→p)^(a→p) = (gp) ^¬(a → p) =F (a) Show that (pq) ^ (q→p) and F are not logically equivalent by finding a counterex- ample. (b) Identify the error(s) in this proof and justify why they are errors. Justify the other steps with their corresponding laws of propositional logic.

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