Practice Exam. 2x 1 + 4x 2 + 2x 3 = 4 x 1 + 2x 2 + 3x 3 = 1 2x 1 + 3x 2 + 4x 3 = 5

Transcription

1 Practice Exam. Solve the linear system using an augmented matrix. State whether the solution is unique, there are no solutions or whether there are infinitely many solutions. If the solution is unique, give it; if there are infinitely many, give the solution parametrically. x + x + x = x + x + x = x + x + x = The augmented matrix of this system has the following form: We proceed as follows to find the RREF of this matrix: (I) (I) Next we switch the last two rows and then: (II) (III) (III) Therefore, x =, x = and x =. In particular, there is a unique solution to this system of linear equations.

2 . Determine all values of s and t such that the following linear system x + x = s x + tx = 8 (a) has no solutions; (b) has a unique solution; (c) has infinitely many solutions; Give separate answers for each part, and justify them. [ } t {{ } A The determinant of the matrix A is equal to: [ x x = [ s 8 det(a) = t = t Therefore, if t, then A is invertible and the above solution can be solved as: [ [ x = A s = [ [ t s = x 8 t 8 [ ts t s+8 t Consequently, if t, then the system has a unique solution for fixed values of s and t. This unique solution is given above. Next, we consider the case that t =. Then thye augmented matrix of this system has the following form: [ s 8 We switch the two rows of this matrix and then proceed as follows to fined the RREF of this matrix: [ [ [ 8 s s (I) s If s, then the system is consistent. If s =, then the above system has infinitely many solutions. In summary, we have the following three cases: (a) t, s R: unique solution, (b) t =, s : no solutions, (c) t =, s = : infinitely many solutions.

3 . The norm of an m n matrix A is defined to be the square root of tr(a t A). (Recall that tr stands for the trace of a square matrix. The trace of a square matrix is the sum of the diagonal terms). Consider the following matrices: A = In each part compute the norm of the given matrix: (a) A B = A t = [ = A t A = [ = [ = (b) B Therefore, the norm of A is equal to. = tr(a t A) = + = B t = [ = B t B = [ = [ 6 = Therefore, the norm of B is equal to 6. (c) A B A B = = (A B) t (A B) = = tr(b t B) = + 6 = 6 = (A B)t = [ Therefore, the norm of A B is equal to. [ = tr((a B) t (A B)) = + = [ = = =

4 . For each of the following matrices decide whether the given matrix is invertible. If it is invertible find the inverse. If not, explain why. (a) 6 (b) (c) (d) No, because it is not a square matrix. 7 By Sarrus rule, the determinant of this matrix is equal to: = Therefore, this matrix is invertible. The inverse of this matrix can be computed by finding the RREF of the following matrix: () 7 This matrix can be simplified to the reduced row echelon form in the following way: (III) 7 (I) (II) (II) 7 (III) (II) Therefore, the inverse is equal to: 7 By Sarrus rule, the determinant of this matrix is equal to: = Therefore, this matrix is not invertible. (Hint: Show that the matrix is orthogonal.)

5 The columns of this matrix form an orthonormal basis because they have length and every two of them are orthogonal to each other. As a result the inverse of this matrix is equal to its transpose:

6 . Find bases for the image and the kernel of the linear transformation defined by the following matrix: A = We firstly find the RREF of A. We firstly switch the first and the third rows: (II) (I) (I) ( ) (II) (I) (II) (IV) (III) +(III) Therefore, a basis for the image of A is given by the first, the second and the fourth columns of A: An element in the kernel of A has the following form: t t t = t Therefore, a basis for the kernel of A is: 6

7 6. Determine whether or not the following vectors are linearly independent. (a) v = v = v = 8 Method I: We have v = v + v. Therefore v is redundant and these vectors are not linearly independent. Method II: Form a matrix whose columns are equal to the vectors v, v and v : (III) (I) 8 (I) (III) (II) Because the last column corresponds to a free variable, it implies that these vectors are not linearly independent. (b) v = v = v = Form a matrix whose columns are equal to the vectors v, v and v : (I) (III) (III) (I) Therefore, these three vectors are linearly independent. +(II) (III) (II) 7

8 7. Following vectors form a basis B = { v, v, v } of R : v = v = v = 6 Let T be the linear transformation which is given by the following matrix: (a) Find the B-matrix of this linear transformation. That is to say, find a matrix B which maps [ x B to [T ( x) B, for each x R. S = 6 We can find the inverse of S similar to the part (b) of #. This inverse is equal to: S = The matrix B is equal to S AS which can be computed as: 8 B = S AS = 7 (b) Find the B-coordinate of the following vector: B = S = 8

9 8. Let A be the linear transformation given by the following matrix: cos θ sinθ sin θ cosθ and θ [,π Let: e = e = e = (a) What are the images of e, e, e under the action of A? cos θ sin(θ) A e = sin θ A e = cos(θ) A e = (b) Show that the action of A preserves the length of vectors, i.e., show that for any vector v R, we have v = A v. As it is shown in the next part, A is an orthogonal matrix. Therefore, it preserves the length of vectors in R. (c) Determine if the images of e, e, e form an orthonormal basis of R. A e A e = cos (θ) + sin (θ) = A e A e = sin (θ) + ( cos(θ)) = A e A e = It is also straightforward to show that the pairwise inner product of these three vectors are equal to. Therefore, the columns of A form an orthonormal basis for R. That implies that A is an orthogonal matrix. 9

10 9. Consider the following two matrices A = B = (a) Determine eigenvalues and the corresponding eigenspaces of A. By Sarrus s rule: det(a λid) = ( λ)( λ)( λ) + ( λ) = ( λ)(( λ)( λ) + ) = ( λ)( λ)( λ) Therefore, the only eigenvalue is equal to. To find the -eigenspace, we need to determine ker(a Id). To that end, we find the RREF of A Id: (III) (I) (I) (III) (III) An element in the kernel of A Id has the following form: t = t (b) Determine whether or not A is diagonalizable, and justify your answer. If A is diagonalizable, find the diagonal matrix (the diagonalized version of A). The previous part shows that the eigenspace for the only eigenvalue of A has dimension (less than ). Therefore, A is not diagonalizable. (c) Determine eigenvalues and the corresponding eigenspaces of B. The matrix B λid is upper-triangular. Therefore, its determinant is equal to the product of its diagonal terms: det(b λid) = ( λ)( λ)( λ) Therefore, the only eigenvalue is equal to. To find the -eigenspace, we need to determine ker(a Id). To that end, we find the RREF of B Id: (I) An element in the kernel of B Id has the following form: t s = t + s

11 (d) Determine whether or not B is diagonalizable, and justify your answer. If B is diagonalizable, find the diagonal matrix (the diagonalized version of B). The previous part shows that the eigenspace for the only eigenvalue of B has dimension (less than ). Therefore, B is not diagonalizable. (e) (Optional) Determine if A is similar to B, and justify your answer. No, because the -eigenspaces for these two matrices do not have the same dimension (cf. Theorem 7.. of the textbook).

12 . Use the Gram-Schmidt process to write an orthonormal basis for the subspace spanned by the vectors below: v = v = v = u = v v = u = v (v u )u v (v u )u = v u v u = = ( ) + ( ) + ( ) + ( ) u = v (v u )u (v u )u v (v u )u (v u )u = v u + u v u + u = ( ) + () + ( ) + () + ( ) = The three vectors {u, u, u } gives an orthonormal basis for the span of v, v and v. =

13 . Consider: v = u = u = 6 6 (a) Show that { u, u } are orthonormal. u u = ( ) ( ) = + = u u = ( ) + ( 6 ) + ( 6 ) + ( ) = = u u = = Therefore, u and u are orthonormal (b) Find the orthogonal projection of v onto span( u, u ). Proj span( u, u ) v = ( v u ) u +( v u ) u = ( +++ ) = (9 + ) 6 +( ) 6 6 =