Name ________________________                         Linear Algebra, Quiz 2, Summer 2004        

In each Problem 1-3 below, (a)  find the determinant of the coefficient matrix A, (b)  if A is invertible, find the inverse of A by row reducing an appropriate augmented matrix and check your answer, (c)  if A is invertible, use your answer in (b)  to solve the system of equations and check your answer. Note that you must show all your work on each problem to get full credit.

Problem 1.

`Problem 1.  System of Equations:`

-2*x+y+z = 0

-4*x+3*y-2*z = -1

2*x-z = -7

`Problem 1.  Matrix Equation:`

A*X = B

matrix([[-2, 1, 1], [-4, 3, -2], [2, 0, -1]])*matrix([[x], [y], [z]]) = matrix([[0], [-1], [-7]])

 

Name ________________________                         Linear Algebra, Quiz 2, Summer 2004        

Problem 2.

`Problem 2.  System of Equations:`

3*x-5*z = -3

x-y = 0

x+2*z = 2

`Problem 2.  Matrix Equation:`

A*X = B

matrix([[3, 0, -5], [1, -1, 0], [1, 0, 2]])*matrix([[x], [y], [z]]) = matrix([[-3], [0], [2]])

 

Name ________________________                         Linear Algebra, Quiz 2, Summer 2004        

Problem 3.

`Problem 3.  System of Equations:`

4*x+2*y-8*z = -7

-2*x+y+4*z = 4

3*x+y-6*z = -3

`Problem 3.  Matrix Equation:`

A*X = B

matrix([[4, 2, -8], [-2, 1, 4], [3, 1, -6]])*matrix([[x], [y], [z]]) = matrix([[-7], [4], [-3]])

 

Name ________________________                         Linear Algebra, Quiz 2, Summer 2004        

Problem 4.

Find a subset of the following five vectors that is a basis for the space generated by these five vectors. Note that you must show (explain) all your work to get full credit.

v[1] = matrix([[1, 1, 3, 2]]), v[2] = matrix([[0, -2, -1, -2]]), v[3] = matrix([[-3, 1, -7, -2]]), v[4] = matrix([[2, 1, 2, 1]]), v[5] = matrix([[6, 5, 7, 5]])

 

Name ________________________                         Linear Algebra, Quiz 2, Summer 2004        

Problem 5.

Without computing anything, find a matrix A satisfying the following equation and completely explain why your answer works.

matrix([[1, 0, 0], [0, -10, 0], [0, 0, 1]])*matrix([[-1, 0, 0], [0, 1, 0], [0, 0, 1]])*matrix([[1, 0, 0], [0, 1, 0], [-4, 0, 1]])*matrix([[1, 0, 0], [0, 1, 3/2], [0, 0, 1]])*matrix([[1, 0, 0], [0, 1, -...