MATH 15a: Applied Linear Algebra Practice Exam 1

Transcription

1 MATH 5a: Applied Linear Algebra Practice Exam Note: this practice test is NOT a guarantee of what the actual midterm will look like!. Say whether the following functions are linear. If so, write down the matrix for T. If not, give an example that shows why not. (a) (b) T x = x T x x x 4 [ ] x +3x 4 x + = x +x x 4x +7x 4. For which values of b and c is the following matrix invertible? 0 b 0 c b c 0 3. Let T : R R be the linear transformation given by first stretching by a factor of 3 in the horizontal direction, then rotating counterclockwise by π/4 radians. (a) Find the matrix that represents T. (b) Using Gauss-Jordan elimination, find the matrix that represents T. (c) Describe T geometrically, and show that this agrees with your answer to (b). 4. (a) Using Gauss-Jordan elimination, find all solutions (if any) to the system of equations x 4x + 5x 4 +6x 5 = 3 x +8x 7 x 4 = 4 x +4x 5 7x 4 +6x 5 = 7 (b) Compute the rank of the matrix True or false? Justify your answer. (a) If the system A x = b has a unique solution, then A must be a square matrix. (b) If V is a subspace of R 3 containing two vectors that are not on the same line, then V equals all of R Let A be an invertible m m matrix, and let B be an m n matrix. Prove that ker(b) = ker(ab). (Hint: You have to show two things: every element of ker(b) is in ker(ab), and every element of ker(ab) is in ker(b).)

2 7. (a) Complete the definition: A set V is called a subspace if... (b) Is the set of vectors x satisfying x +x + = 0 a subspace of R 3? x

3 MATH 5a: Applied Linear Algebra Practice Exam, Solutions Note: this practice test is NOT a guarantee of what the actual midterm will look like!. Say whether the following functions are linear. If so, write down the matrix for T. If not, give an example that shows why not. (a) T x x = [ ] x +3x 4 x + Answer: Any linear function must have T( 0) = 0, but this function has 0 [ 0 0 = ] 0 (b) T x x x 4 = x +x x 4x +7x 4 Answer: It is linear, and the matrix is For which values of b and c is the following matrix invertible? 0 b 0 c b c 0 Answer: We can do row-reduction: 0 b 0 c 0 c 0 b 0 c 0 b 0 c 0 b b c 0 b c 0 0 c bc 0 0 bc+bc Since the last row is all zeroes, this matrix is never invertible, no matter what b and c are. 3. Let T : R R be the linear transformation given by first stretching by a factor of 3 in the horizontal direction, then rotating counterclockwise by π/4 radians. (a) Find the matrix that represents T. Answer: The point (,0) goes to (3,0) under the stretch, and then to ( 3, 3 ) under therotation. Thepoint(0,)staysfixedunderthestretchandthengoesto( under the rotation. Hence, the matrix is [ 3 3 ], )

4 (b) Using Gauss-Jordan elimination, find the matrix that represents T. Answer: We ll cheat a little and use the formula [ ] [ ] a b d b =. c d ad bc c a (You should know this formula for the test.) The answer is thus ] [ 6 6 (c) Describe T geometrically, and show that this agrees with your answer to (b). Answer: Following the steps in reverse, T rotates clockwise by π/4 and then stretches horizontally by a factor of 3. Thus, the point (,0) goes to ( 6, ), and the point (0,) goes ( ), as above. 6, 4. (a) Using Gauss-Jordan elimination, find all solutions (if any) to the system of equations x 4x + 5x 4 +6x 5 = 3 x +8x 7 x 4 = 4 x +4x 5 7x 4 +6x 5 = 7 Answer: Starting with the augmented matrix and using Gauss-Jordan elimination, we have: = = / = / / The pivot variables are x and, and the free variables are x, x 4, and x 5. Thus: x = 4x +3x 4 +x 5 /3 x = free = 4x 4 +4x 5 +0/3 x 4 = free x 5 = free (b) Compute the rank of the matrix Answer: We saw in part (a) that the RREF of this matrix is which has two pivots. Thus the rank is.

5 5. True or false? Justify your answer. (a) If the system A x = b has a unique solution, then A must be a square matrix. Answer: False. It s easy to come up with systems with more equations than variables that have unique solutions; of course, some of the equations will be redundant. The system x = x = is an example. The matrix for this system is, hence not square. (b) If V is a subspace of R 3 containing two vectors that are not on the same line, then V equals all of R 3. Answer: False. The xy plane contains the vectors e and e, but it is not all of R Let A be an invertible m m matrix, and let B be an m n matrix. Prove that ker(b) = ker(ab). (Hint: You have to show two things: every element of ker(b) is in ker(ab), and every element of ker(ab) is in ker(b).) Answer: Forthefirststep, if v ker(b), thenb v = 0bydefinition. Hence(AB) v = A(B v) = A 0 = 0. Thus, v ker(ab). For the second (and harder) step, if v ker(ab), then (AB) v = 0. Since A is invertible, we can multiply both sides of this equation on the left by A to get: Therefore, v ker(b). A AB v = A 0 (A A)B v = 0 I m B v = 0 B v = 0 7. (a) Complete the definition: A set V is called a subspace if... Answer:...it satisfies the following three properties: i. 0 is in V; ii. If v and w are in V, then v + w is in V; iii. If v is in V, then c v is in V for any scalar c. (b) Is the set of vectors x satisfying x +x + = 0 a subspace of R 3? x Answer: Yes. You can recognize that set as the kernel of the matrix [ ], and the kernel of any matrix is a subspace. Alternately, we can check each of the properties explicitly: i. 0 is inv because = 0. ii. If v = x and w = y areboth in V, then x +x + = 0 and y +y +y 3 = x y Y 3 0. Hence (x + y ) + (x + y ) + ( + y 3 ) = 0, which means that the vector v + w = x +y x +y is also in V. +y 3 3

6 iii. If v = x x is in V, then cx +cx +c = 0, so c v = cx cx is in V. c 4