# University of Lille I PC first year list of exercises n 7. Review

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1 University of Lille I PC first year list of exercises n 7 Review Exercise Solve the following systems in 4 different ways (by substitution, by the Gauss method, by inverting the matrix of coefficients of the system, by Cramer s formulas) : { x + y = 3x + 7y = Choose the method that seems the quickest to you and solve, according to the values of a, the following systems : { ax + y = (a + )x + ay = Solution of Exercise : { (a + )x + (a )y = (a )x + (a + )y = (a) By substitution { { { x + y = x + y = 4 3x + 7y = x = 7 3 y 3 y + y = x = 7 3 y { y = 3 x = 7 (b) By the Gauss method { { x + y = x + y = 3x + 7y = y = 3 L L 3L { x = y y = 3 { (c) The inverse of the matrix of coefficients of the system is ( 3 7 ) = ( 7 3 y = 3 x = 7 ) Hence the solution of the system is ( ) x y = ( 7 3 ) ( ) = ( 7 3 ) (d) By Cramer s formulas x = = 7 y = = 3

2 The determinant of the first system is a (a + ) a = a (a) If a / {, }, one can use Cramer s formulas to obtain : a x = { ax + y = (a + )x + ay = y = a (a + ) a a (a + ) a (a + ) a = 4a a = a + a a (b) If a =, the system becomes { { x + y = x + y = x + y = = L L L which is impossible (c) If a =, the system becomes { { x + y = x + y = x y = = 5 L L + L which is also impossible The determinant of the second system is (a + ) (a ) (a ) (a + ) = 4a (a) If a, one can use Cramer s formulas to obtain : { (a + )x + (a )y = (a )x + (a + )y = x = y = (a ) (a + ) 4a (a + ) (a ) (b) If a =, the system becomes { { x y = x y = x + y = = L L + L which is impossible Exercise Solve the following system of 5 equations with 6 unknowns : x + y + z u + 3v w = 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + v w = 3 3x 3u + 3v + 3w = 6 4a = a = a

3 Solution of Exercise : By the Gauss method x + y + z u + 3v w = 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + v w = 3 3x 3u + 3v + 3w = 6 x + y + z u + v w = 3 L L 4 3x + y + z 3u + 5v 3w = 4 x + y + z u + 4v 4w = 6 x + y + z u + 3v w = 3x 3u + 3v + 3w = 6 x + y + z u + v w = 3 y z v + 3w = 5 L L 3L = L 3 L 3 L y z v + 3w = 5 L 4 L 4 L 3y 3z 3v + 9w = 5 L 5 L 5 3L { x + y + z u + v w = 3 y z v + 3w = 5 It follows that the set of solutions is a 4-space in R 6 Let us parametrize the set of solutions by a = z R, b = u R, c = v R, d = w R One obtains x y z u v w = x = y a + b c + d + 3 = b c d y = a c + 3d + 5 z = a u = b v = c w = d 5 + a + b Exercise 3 For each pair (A i, b i ), i 5 of matrices below + c give the nature of the set of solutions of the system A i X = b i ; give a parametric representation of the set of solutions of A i X = b i ; 3 give a basis of the range and a basis of the kernel of A i + d 3 3

4 a) A = c) A 3 = e) A 5 = b = Solution of Exercise 3 : b 3 = b 5 = ; b) A = ; d) A 4 = ; 3 3 b = b 4 = a) Since deta =, the matrix A is invertible hence defines an isomorphism of R 4 The system A X = b has therefore a unique solution given by X = A b = (,,, ) T by a standard computation The range of A is R 4, hence the canonical basis of R 4 is a basis of Im A The kernel of A is { }, hence a basis of kera is b) The rank of A is 4, hence the dimension of the kernel of A is Therefore the set of solutions of A X = b is an affine line in R 5 parallel to kera Denote by (x, y, z, t, u) the coordinates in R 5 Let us parametrize the set of solutions by a = u R The system is equivalent to x + y + t = 3a y + z + t = a z + t = 3a t = a x = 3a + a = a y = a + + a + ( a) = z = 3a + a = a t = a x = 3a y t y = a z t z = 3a t t = a x y z t u = + a ; ;, a R Since A is surjective, the canonical basis of R 4 is a basis of ImA The previous resolution implies that a basis of kera is given by the single vector c) Since the last equation of the system is impossible, the system A 3 X = b 3 admits no solution The rank of A 3 is 4, therefore by the Rank theorem, the dimension of kera 3 is A basis of ImA 3 is given by the 4 columns of A 3 A basis of kera 3 is given by the empty set d) The last equation of A 4 X = b 4 is impossible, hence this system admits no solution The rank of A 4 is 4, hence by the Rank theorem, the dimension of the kernel of A 4 is A basis of ImA 4 is given by the first 4 columns of A 4 A basis of kera 4 is a nontrivial vector X R 5 solution of A 4 X = One finds that generates kera 4 4

5 e) For the basis of ImA 5 and kera 5 see d) The vector b 5 belongs to ImA 5 since the last equation (compatibility condition) is satisfied The kernel of A 5 being a line, the set of solutions of A 5 X = b 5 is an affine line in R 5 parallel to kera 5 Since the vector is a particular solution of the system, one obtains that the set of solutions is parametrized by x y z t = + a, a R u Exercise 4 Compute a basis of the image and a basis of the kernel of the linear application What is the rank of f? f : R 3 R 5 (x, y, z) (x + y, x + y + z, x + y + z, x + y + z, y + z) Solution of Exercise 4 : The matrix of the linear application f is Let us compute a basis of Imf and a basis of kerf One has : Consequently the kernel of f is trivial, and a basis of Imf is given by v = v 3 = The rank of f is the dimension of Imf, that is, 3 Exercise 5 Let A be the matrix 3, v = and 5

6 Consider the matrices B = the matrix A be invertible? and C = Show that AB = AC Can Determine all matrices F of size (3, 3) such that AF = (where denotes the matrix all of whose entries are zero) Solution of Exercise 5 : One has AB = AC = Suppose that the matrix A is invertible Multiply both members of the equation AB = AC on the left by A to get B = C But the matrices B and C are not equal This is a contradiction Hence the matrix A is not invertible Let F be any real matrix (3, 3) F = a b c d e f g h i The equation AF = gives rise to the following system a = b = c = d + g = e + h = f + i = 3a + d + g = 3b + e + h = 3c + f + i = Consequently the set of matrices F such that AF = is the set of matrices of the form F = d e f, d R, e R, f R d e f Exercise 6 For which values of a is the matrix invertible? Compute in this case its inverse A = 4 3 a Solution of Exercise 6 : One has det A = 4 3 a = 4 3 a 3 a + 4 = a (a 3) + = a 7 Hence A is invertible if and only if a 7 In this case, the standard algorithm yields A = a 3 a 4 a a 3 a 7 6

7 Exercise 7 Let a and b be two real numbers, and A be the matrix a b A = Show that rk(a) (where rk denotes the rank) For which values of a and b is the rank of A equal to? Solution of Exercise 7 : Recall that the rank of A is the greatest number of columns of A that are linearly independent Since the second and third columns C, C 3 of A are not proportional, they are linearly independent Therefore the rank of A is at least For the rank of A to be exactly, one has to impose that the first and last columns of A are each a linear combination of C and C 3 (which are fixed) The only linear combination of C and C 3 that has the form (a, 3, 5) T is 3C 3 + C = (, 3, 5) T, hence a = The only linear combination of C and C 3 that has the form (b, 4, ) T is 4C 3 C = (3, 4, ) T, hence b = 3 Consequently the rank of A is if and only if a = and b = 3 Exercise 8 Compute the inverse of the following matrix Solution of Exercise 8 : One obtains A = A = Exercise 9 Let us denote by {e, e,, e n } the canonical basis of R n To a permutation σ S n, one associates the following endomorphism u σ of R n : u σ : R n R n x x n Let τ = (ij) be a transposition Write the matrix of u τ in the canonical basis Show that det(u τ ) = Show that σ, σ S n, u σ u σ = u σ σ Caution! There was a typo in the French original x σ() x σ(n) 3 Show that σ S n, det u σ = ε(σ) where ε denotes the signature Solution of Exercise 9 : 7

8 Let τ be the transposition which exchanges i and j The matrix of u τ in the canonical basis of R n is i j By exchanging the columns i and j of the matrix of u τ one obtains the identity matrix Therefore det u τ = det I =, where I denotes the identity matrix For any σ, σ S n, one has u σ u σ x x n = u σ x σ () x σ (n) = Since the previous equality is satisfied for every x σ (σ()) x σ (σ(n)) x x n = x σ σ() x σ σ(n) = u σ σ x x n in R n, it implies that u σ u σ = u σ σ An alternative proof is to check that u σ sends each e i to e σ (i) (the basis vector whose only nonzero coordinate is the σ (i)-th) : hence, u σ u σ (e i ) = u σ (e σ (i) ) = e σ (σ (i)) = e (σ σ) (i) = u σ σ(i) 3 By, the map which associates u σ to a permutation σ is a group homomorphism from S n into the group of invertible matrices of size (n, n), because u σ u σ = u σ σ = u (σ σ ) Consequently, the map which assigns to a permutation σ the number det u σ is a group homomorphism from S n into {±} Since the transpositions generate the group of permutations S n, two group homomorphisms from S n to {±} which coincide on the set of transpositions coincide on S n By, the group homomorphism from S n into {±} which maps σ onto det u σ coincides with the signature on the set of transpositions, because a transposition is its own inverse Hence σ S n, det u σ = ε(σ) Exercise Compute the eigenvalues and eigenvectors of the following matrix A = 3 4 Compute A n for all n N Solution of Exercise : 8

9 One has det (A λi) = λ λ 3 4 λ = λ λ 3 4 λ 3 4 λ + λ = λ 3 + 3λ λ = λ (λ ) (λ ) Therefore the eigenvalues of A are λ =, λ = and λ 3 = A nontrivial vector in the kernel of A is given by v = Let us find a vector generating the eigenspace associated to λ = One has A λ I I = 3 3 It follows that the vector v = has Consequently the vector v 3 = C C + C C 3 C 3 C C 3 C 3 + C is a basis of the eigenspace associated to λ = Now one A I = 3 3 generates the eigenspace associated to λ 3 = Denote by f the linear application whose matrix in the canonical basis of R 3 is A The vectors v, v and v 3 form a basis of R 3 In this new basis, the matrix of f is D = The relation between A and D is D = P AP where P = P = The inverse of P is Therefore, for n >, we have A n = (P DP )(P DP ) (P DP ) : cancelling all occurrences of P P = I one gets n n n A n = P D n P = n = n + n n n+ + n+ 9

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