MAT 242 Test 2 SOLUTIONS, FORM T

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1 MAT 242 Test 2 SOLUTIONS, FORM T Let v =, v 5 2 =, v 3 =, and v 5 4 = a. [ points] The set { v, v 2, v 3, v 4 } is linearly dependent. Find a nontrivial linear combination of these vectors that adds up to. To do this, you need to find a nontrivial solution to c v + c 2 v 2 + c 3 v 3 + c 4 v 4 =, whose matrix looks like [ v v 2 v 3 v 4 ], or RREF The nontrivial solutions are thus given by c = 2t c 2 = 5t c 3 = t c 4 = where t can be any real number Choosing t = produces the linear combination 2v 5v 2 + v 3. Grading: +5 points for setting up the system of linear equations; +5 points for parameterizing it; +5 points for finding a linear combination. Grading for common mistakes: +7 points (total) for using the column space method instead. 4 3 b. [ points] Is the vector in the span of { v 6, v 2, v 3, v 4 }? Justify your answer. 5 You must determine whether the augmented matrix [ v v 2 v 3 v 4 u] is a system that has at least one solution RREF Since this system has no solutions, the vector is not in the span. Grading: +4 points for setting up the matrix, +4 points for the RREF, +3 points for determining how many solutions there were, +4 points for answering NO. Grading for common mistakes: 8 points for using instead of u. 2

2 MAT 242 Test 2 SOLUTIONS, FORM T 2. [25 points] For the matrix A below, find a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A, and the nullity of A. The reduced row echelon form of A is the matrix R given below A = R = To find a basis for the null space, you need to solve the system of linear equations A x =, or equivalently, R x =. Parameterizing the solutions to this equation, and writing the answer in expanded form, produces x x 3 = α + α 2 so x 4 x , is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R: {[,, 5, 2, ], [,, 3,, ], [,,,, ]} A basis for the column space can be found by taking the columns of A which have pivots in them, so ,, is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space) of A, so the rank of A is 3. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a basis for the column space, the nullity, and the rank. Grading for common mistakes: 3 points for forgetting a variable in the parameterization; 3 points for choosing columns of R for the column space of A; 3 points for choosing rows from A for the row space of A; 3 points for choosing the non-pivot columns of A for the null space of A. 3

3 MAT 242 Test 2 SOLUTIONS, FORM T [5 points] Find the eigenvalues of A = Solve for when the determinant of A λi equals zero: 6 λ λ λ = λ3 + λ 2 6λ = λ (λ + 3) (λ 2). The eigenvalues make this product equal to zero, so they are, 3, and 2. Grading: +5 points for A λi, +5 points for finding the determinant, +5 points for finding the eigenvalues. Grading for common mistakes: +5 points (total) for using row operations on the matrix A, +3 points (total) for finding the eigenvectors of. 4. Let B be the (ordered) basis 2, 2 5, 3 9 and C the basis, 3, a. [ points] Find the coordinates of 6 8 with respect to the basis C. 38 [ u] C = C u = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation. b. [ points] Find the change-of-basis matrix from C to B. 2 3 B C = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation Grading for common mistakes: +7 points (total) for C B = (backwards), +5 points (total) for BC = 2 2, +5 points (total) for CB = 3 2 4, +3 points (total) for B = , +7 points (total) for B C x 396 6, +3 points (total) for any other 282 vector. 4

4 MAT 242 Test 2 SOLUTIONS, FORM T 5. The eigenvalues of the matrix A = are (with multiplicity 2) and 2 (with multiplicity ). 2 2 (You do not need to find these.) Do the following for the matrix A: a. [ points] Find a basis for the eigenspace of each eigenvalue. The eigenspace of an eigenvalue λ is the null space of A λi. So, if λ =, A ()I = RREF, 2 2 whose solutions (with a vector of s on the right-hand side) are parameterized by and a basis for the eigenspace of is If λ = 2, whose solutions are parameterized by and a basis for the eigenspace of 2 is x = s + t x 3,. A (2)I = RREF 3 2, 2 4 x = s 3 2 x Grading: +3 points for A λi, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A lambdai. b. [ points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = P DP and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not. YES. The dimension of each eigenspace equals the (given) multiplicity of each eigenvalue. The matrix P can be constructed by gluing the basis vectors together; then the (i, i) entry in D is the eigenvalue that the ith column of P belongs to. One pair of matrices that diagonalizes A is D = and P = Grading: +3 points for YES, +7 points for P and D. Full credit marked + points was given for an answer consistent with any mistakes made in part (a). 5

5 MAT 242 Test 2 SOLUTIONS, FORM W Let S =,,,, a. [ points] Find a basis for the subspace W spanned by S, and the dimension of W. To find a basis, use the column space (CS) approach: Glue the vectors together to get the matrix Ṽ, find the RREF, and take the original vectors that have pivots in their columns: Ṽ = 2 RREF A basis for the subspace will consist of the columns of Ṽ which have a pivot in their column, namely ,,. The dimension of the subspace is the number of vectors in this basis, which is 3. Grading: +5 points for Ṽ, +5 points for the RREF, +5 points for the dimension. Grading for common mistakes: +5 points (total) for finding a basis for the null space; 3 points for using the columns of the RREF. 7 b. [ points] Is the vector in the span of S? Justify your answer. 4 3 You must determine whether the augmented matrix [ v v 2 v 3 v 4 u] is a system that has at least one solution RREF Since this system has infinitely many solutions, it has at least one, and the vector is in the span. Grading: +4 points for setting up the matrix, +4 points for the RREF, +3 points for determining how many solutions there were, +4 points for answering YES. Grading for common mistakes: 8 points for using instead of u. 2

6 MAT 242 Test 2 SOLUTIONS, FORM W 2. [25 points] For the matrix A below, find a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A, and the nullity of A. The reduced row echelon form of A is the matrix R given below A = R = To find a basis for the null space, you need to solve the system of linear equations A x =, or equivalently, R x =. Parameterizing the solutions to this equation, and writing the answer in expanded form, produces 5 so x 3 = α x + α 3 2 x , 5 is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R: {[,,, 5 ], [,, 3, 5 ]} A basis for the column space can be found by taking the columns of A which have pivots in them, so , is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space) of A, so the rank of A is 2. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a basis for the column space, the nullity, and the rank. Grading for common mistakes: 3 points for forgetting a variable in the parameterization; 3 points for choosing columns of R for the column space of A; 3 points for choosing rows from A for the row space of A; 3 points for choosing the non-pivot columns of A for the null space of A. 3

7 MAT 242 Test 2 SOLUTIONS, FORM W 3. [5 points] Find the eigenvalues of A = Solve for when the determinant of A λi equals zero: λ 3 3 λ 3 λ = λ3 + 3λ 2 = λ 2 (λ + 3). The eigenvalues make this product equal to zero, so they are,, and 3. Grading: +5 points for A λi, +5 points for finding the determinant, +5 points for finding the eigenvalues. Grading for common mistakes: +5 points (total) for using row operations on the matrix A, +3 points (total) for finding the eigenvectors of. 4. Let B be the (ordered) basis 3, 2 7, 2 and C the basis, 3, a. [ points] Find the coordinates of 9 with respect to the basis B. 23 [ u] B = B 2 u = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation. b. [ points] If the coordinates of u with respect to B are 6, what are the coordinates of u with 2 respect to C? [ u] C = C B [ u] B = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation. Grading for common mistakes: +7 points (total) for B C[ u] = (backwards), +5 points 9 (total) for CB [ u] = , +5 points (total) for BC [ u] = 7, +3 points (total) for C [ u] = , +7 points (total) for the change-of-basis matrix C B = 3 7 2, points (total) for a different change-of-basis matrix. 4

8 MAT 242 Test 2 SOLUTIONS, FORM W 5. The eigenvalues of the matrix A = 3 3 are 3 (with multiplicity 2) and 4 (with multiplicity ). (You do not need to find these.) Do the following for the matrix 4 A: a. [ points] Find a basis for the eigenspace of each eigenvalue. The eigenspace of an eigenvalue λ is the null space of A λi. So, if λ = 3, A ( 3)I = RREF, whose solutions (with a vector of s on the right-hand side) are parameterized by and a basis for the eigenspace of 3 is If λ = 4, whose solutions are parameterized by and a basis for the eigenspace of 4 is x = s + t x 3,. A ( 4)I = RREF, x = s x 3.. Grading: +3 points for A λi, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A lambdai. b. [ points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = P DP and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not. YES. The dimension of each eigenspace equals the (given) multiplicity of each eigenvalue. The matrix P can be constructed by gluing the basis vectors together; then the (i, i) entry in D is the eigenvalue that the ith column of P belongs to. One pair of matrices that diagonalizes A is D = 3 3 and P = 4 Grading: +3 points for YES, +7 points for P and D. Full credit marked + points was given for an answer consistent with any mistakes made in part (a). 5

9 MAT 242 Test 2 SOLUTIONS, FORM O Let S =,,,, a. [ points] Find a basis for the subspace W spanned by S, and the dimension of W. To find a basis, use the column space (CS) approach: Glue the vectors together to get the matrix Ṽ, find the RREF, and take the original vectors that have pivots in their columns: Ṽ = 4 2 RREF A basis for the subspace will consist of the columns of Ṽ which have a pivot in their column, namely 3 4 3,,. The dimension of the subspace is the number of vectors in this basis, which is 3. Grading: +5 points for Ṽ, +5 points for the RREF, +5 points for the dimension. Grading for common mistakes: +5 points (total) for finding a basis for the null space; 3 points for using the columns of the RREF. 65 b. [ points] Is the vector in the span of S? Justify your answer You must determine whether the augmented matrix [ v v 2 v 3 v 4 u] is a system that has at least one solution RREF Since this system has no solutions, the vector is not in the span. Grading: +4 points for setting up the matrix, +4 points for the RREF, +3 points for determining how many solutions there were, +4 points for answering NO. Grading for common mistakes: 8 points for using instead of u. 2

10 MAT 242 Test 2 SOLUTIONS, FORM O 2. [25 points] For the matrix A below, find a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A, and the nullity of A. The reduced row echelon form of A is the matrix R given below A = R = To find a basis for the null space, you need to solve the system of linear equations A x =, or equivalently, R x =. Parameterizing the solutions to this equation, and writing the answer in expanded form, produces x x 3 = α + α 2 so x 4 x , is a basis for the null space of A. The nullity is the number of vectors in this basis, namely 2. A basis for the row space can be found by taking the nonzero rows of R: {[,,, 4, 3 ], [,,,, 5 ], [,,,, ]} A basis for the column space can be found by taking the columns of A which have pivots in them, so , 4, is a basis for the column space of A. Lastly, the rank of A is the number of vectors in a basis for the row space (or column space) of A, so the rank of A is 3. Grading: +5 points for each of: finding a basis for the null space, a basis for the row space, a basis for the column space, the nullity, and the rank. Grading for common mistakes: 3 points for forgetting a variable in the parameterization; 3 points for choosing columns of R for the column space of A; 3 points for choosing rows from A for the row space of A; 3 points for choosing the non-pivot columns of A for the null space of A. 3

11 MAT 242 Test 2 SOLUTIONS, FORM O 3. [5 points] Find the eigenvalues of A = Solve for when the determinant of A λi equals zero: λ 4 2 λ 2 2 λ = λ3 4λ = λ (λ + 2) (λ 2). The eigenvalues make this product equal to zero, so they are, 2, and 2. Grading: +5 points for A λi, +5 points for finding the determinant, +5 points for finding the eigenvalues. Grading for common mistakes: +5 points (total) for using row operations on the matrix A, +3 points (total) for finding the eigenvectors of. 4. Let B be the (ordered) basis,, 2 3 and C the basis 3, 2, a. [ points] Find the coordinates of 3 38 with respect to the basis C. 8 [ u] C = C u = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation. b. [ points] If the coordinates of u with respect to B are 8, what are the coordinates of u with respect to C? [ u] C = C B [ u] B = = Grading: +3 points for the formula, +4 points for substitution, +3 points for calculation. Grading for common mistakes: +7 points (total) for B C[ u] = (backwards), +5 points 6 (total) for CB [ u] = , +5 points (total) for BC [ u] = 2, +3 points (total) for C [ u] = , +7 points (total) for the change-of-basis matrix C B = 2 7, points (total) for a different change-of-basis matrix. 4

12 MAT 242 Test 2 SOLUTIONS, FORM O The eigenvalues of the matrix A = 3 6 are 5 (with multiplicity 2) and 3 (with multiplicity ). (You do not need to find these.) Do the following for the matrix 8 A: a. [ points] Find a basis for the eigenspace of each eigenvalue. The eigenspace of an eigenvalue λ is the null space of A λi. So, if λ = 5, 6 32 A ( 5)I = 8 6 RREF 2, 8 6 whose solutions (with a vector of s on the right-hand side) are parameterized by and a basis for the eigenspace of 5 is If λ = 3, whose solutions are parameterized by and a basis for the eigenspace of 3 is x = s + t 2 x 3, A (3)I = 6 6 RREF 2, 8 8 x = s 2 x Grading: +3 points for A λi, +3 points for the RREF, +4 points for finding the null space basis. Grading for common mistakes: +5 points (total) for finding the column space of A lambdai. b. [ points] Is the matrix A diagonalizable? If so, find matrices D and P such that A = P DP and D is a diagonal matrix. If A is not diagonalizable explain carefully why it is not. YES. The dimension of each eigenspace equals the (given) multiplicity of each eigenvalue. The matrix P can be constructed by gluing the basis vectors together; then the (i, i) entry in D is the eigenvalue that the ith column of P belongs to. One pair of matrices that diagonalizes A is D = 5 5 and P = Grading: +3 points for YES, +7 points for P and D. Full credit marked + points was given for an answer consistent with any mistakes made in part (a). 5

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