Name ______Answers___________                         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]])

det(A) = -8, ` so `*A^`-1 `*`exists.`

`Problem 1.  Augment A Using the Identity Matrix to Compute Matrix Inverse:`

A = matrix([[-2, 1, 1], [-4, 3, -2], [2, 0, -1]])*`  I =`*matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

`Augmented matrix:  `*A[0] = matrix([[-2, 1, 1, 1, 0, 0], [-4, 3, -2, 0, 1, 0], [2, 0, -1, 0, 0, 1]])

``

`Swap row1 & row3: `*matrix([[-2, 1, 1, 1, 0, 0], [-4, 3, -2, 0, 1, 0], [2, 0, -1, 0, 0, 1]])*` -->> `*matrix([[2, 0, -1, 0, 0, 1], [-4, 3, -2, 0, 1, 0], [-2, 1, 1, 1, 0, 0]])

`Row reduce using (2),(1): `*matrix([[2, 0, -1, 0, 0, 1], [-4, 3, -2, 0, 1, 0], [-2, 1, 1, 1, 0, 0]])*` -->> `*matrix([[2, 0, -1, 0, 0, 1], [0, 3, -4, 0, 1, 2], [0, 1, 0, 1, 0, 1]])

`Row reduce using (1/2): `*matrix([[2, 0, -1, 0, 0, 1], [0, 3, -4, 0, 1, 2], [0, 1, 0, 1, 0, 1]])*` -->> `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 3, -4, 0, 1, 2], [0, 1, 0, 1, 0, 1]])

`Swap row2 & row3: `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 3, -4, 0, 1, 2], [0, 1, 0, 1, 0, 1]])*` -->> `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 3, -4, 0, 1, 2]])

`Row reduce using (-3): `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 3, -4, 0, 1, 2]])*` -->> `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 0, -4, -3, 1, -1]])

`Row reduce using (-1/4): `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 0, -4, -3, 1, -1]])*` -->> `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 0, 1, 3/4, -1/4, 1/4]])

`Row reduce using (1/2): `*matrix([[1, 0, -1/2, 0, 0, 1/2], [0, 1, 0, 1, 0, 1], [0, 0, 1, 3/4, -1/4, 1/4]])*` -->> `*matrix([[1, 0, 0, 3/8, -1/8, 5/8], [0, 1, 0, 1, 0, 1], [0, 0, 1, 3/4, -1/4, 1/4]])

`Row Reduction:  `*matrix([[-2, 1, 1, 1, 0, 0], [-4, 3, -2, 0, 1, 0], [2, 0, -1, 0, 0, 1]])*` -->>  ...  -->>`*matrix([[1, 0, 0, 3/8, -1/8, 5/8], [0, 1, 0, 1, 0, 1], [0, 0, 1, 3/4, -1/4, 1/4]])

`Inverse of A:  `*A^`-1 ` = matrix([[3/8, -1/8, 5/8], [1, 0, 1], [3/4, -1/4, 1/4]])

`Check:  `*A^`-1`*A = matrix([[3/8, -1/8, 5/8], [1, 0, 1], [3/4, -1/4, 1/4]])*matrix([[-2, 1, 1], [-4, 3, -2], [2, 0, -1]])*` = `*matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

`Solution Vector:  `*X = A^`-1`*B

`Solution Vector:  `*matrix([[x], [y], [z]]) = matrix([[3/8, -1/8, 5/8], [1, 0, 1], [3/4, -1/4, 1/4]])*matrix([[0], [-1], [-7]])

`Solution Vector:  `*matrix([[x], [y], [z]]) = matrix([[-17/4], [-7], [-3/2]])

`Check Matrix Equation: `*matrix([[-2, 1, 1], [-4, 3, -2], [2, 0, -1]])*matrix([[-17/4], [-7], [-3/2]]) = matrix([[0], [-1], [-7]])

`Check Matrix Equation: `*matrix([[0], [-1], [-7]]) = matrix([[0], [-1], [-7]])

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]])

det(A) = -11, ` so `*A^`-1 `*`exists.`

`Problem 2.  Augment A Using the Identity Matrix to Compute Matrix Inverse:`

A = matrix([[3, 0, -5], [1, -1, 0], [1, 0, 2]])*`  I =`*matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

`Augmented matrix:  `*A[0] = matrix([[3, 0, -5, 1, 0, 0], [1, -1, 0, 0, 1, 0], [1, 0, 2, 0, 0, 1]])

``

`Swap row1 & row3: `*matrix([[3, 0, -5, 1, 0, 0], [1, -1, 0, 0, 1, 0], [1, 0, 2, 0, 0, 1]])*` -->> `*matrix([[1, 0, 2, 0, 0, 1], [1, -1, 0, 0, 1, 0], [3, 0, -5, 1, 0, 0]])

`Row reduce using (-1),(-3): `*matrix([[1, 0, 2, 0, 0, 1], [1, -1, 0, 0, 1, 0], [3, 0, -5, 1, 0, 0]])*` -->> `*matrix([[1, 0, 2, 0, 0, 1], [0, -1, -2, 0, 1, -1], [0, 0, -11, 1, 0, -3]])

`Row reduce using (-1): `*matrix([[1, 0, 2, 0, 0, 1], [0, -1, -2, 0, 1, -1], [0, 0, -11, 1, 0, -3]])*` -->> `*matrix([[1, 0, 2, 0, 0, 1], [0, 1, 2, 0, -1, 1], [0, 0, -11, 1, 0, -3]])

`Row reduce using (-1/11): `*matrix([[1, 0, 2, 0, 0, 1], [0, 1, 2, 0, -1, 1], [0, 0, -11, 1, 0, -3]])*` -->> `*matrix([[1, 0, 2, 0, 0, 1], [0, 1, 2, 0, -1, 1], [0, 0, 1, -1/11, 0, 3/11]])

`Row reduce using (-2),(-2): `*matrix([[1, 0, 2, 0, 0, 1], [0, 1, 2, 0, -1, 1], [0, 0, 1, -1/11, 0, 3/11]])*` -->> `*matrix([[1, 0, 0, 2/11, 0, 5/11], [0, 1, 0, 2/11, -1, 5/11], [0, 0, 1, -1/11, 0, 3/1...

`Row Reduction:  `*matrix([[3, 0, -5, 1, 0, 0], [1, -1, 0, 0, 1, 0], [1, 0, 2, 0, 0, 1]])*` -->>  ...  -->>`*matrix([[1, 0, 0, 2/11, 0, 5/11], [0, 1, 0, 2/11, -1, 5/11], [0, 0, 1, -1/11, 0, 3/11]])

`Inverse of A:  `*A^`-1 ` = matrix([[2/11, 0, 5/11], [2/11, -1, 5/11], [-1/11, 0, 3/11]])

`Check:  `*A^`-1`*A = matrix([[2/11, 0, 5/11], [2/11, -1, 5/11], [-1/11, 0, 3/11]])*matrix([[3, 0, -5], [1, -1, 0], [1, 0, 2]])*` = `*matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

`Solution Vector:  `*X = A^`-1`*B

`Solution Vector:  `*matrix([[x], [y], [z]]) = matrix([[2/11, 0, 5/11], [2/11, -1, 5/11], [-1/11, 0, 3/11]])*matrix([[-3], [0], [2]])

`Solution Vector:  `*matrix([[x], [y], [z]]) = matrix([[4/11], [4/11], [9/11]])

`Check Matrix Equation: `*matrix([[3, 0, -5], [1, -1, 0], [1, 0, 2]])*matrix([[4/11], [4/11], [9/11]]) = matrix([[-3], [0], [2]])

`Check Matrix Equation: `*matrix([[-3], [0], [2]]) = matrix([[-3], [0], [2]])

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]])

det(A) = 0, ` so `*A^`-1 `*`does not exists.`

`Problem 3.  Augment A Using the Identity Matrix to Compute Matrix Inverse:`

A = matrix([[4, 2, -8], [-2, 1, 4], [3, 1, -6]])*`  I =`*matrix([[1, 0, 0], [0, 1, 0], [0, 0, 1]])

`Augmented matrix:  `*A[0] = matrix([[4, 2, -8, 1, 0, 0], [-2, 1, 4, 0, 1, 0], [3, 1, -6, 0, 0, 1]])

``

`Row Reduction:  `*matrix([[4, 2, -8, 1, 0, 0], [-2, 1, 4, 0, 1, 0], [3, 1, -6, 0, 0, 1]])*` -->>  ...  -->>`*matrix([[1, 0, -2, 0, -1/5, 1/5], [0, 1, 0, 0, 3/5, 2/5], [0, 0, 0, 1, -2/5, -8/5]])

`A second verification that `*A^`-1 `*`does not exists.`

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]])

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

A^t = matrix([[1, 0, -3, 2, 6], [1, -2, 1, 1, 5], [3, -1, -7, 2, 7], [2, -2, -2, 1, 5]])

`Row reduce using (-1),(-3),(-2): `*matrix([[1, 0, -3, 2, 6], [1, -2, 1, 1, 5], [3, -1, -7, 2, 7], [2, -2, -2, 1, 5]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, -2, 4, -1, -1], [0, -1, 2, -4, -11], [0, -2...

`Swap row2 & row3: `*matrix([[1, 0, -3, 2, 6], [0, -2, 4, -1, -1], [0, -1, 2, -4, -11], [0, -2, 4, -3, -7]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, -1, 2, -4, -11], [0, -2, 4, -1, -1], [0, -2, 4, -3, -...

`Row reduce using (-1): `*matrix([[1, 0, -3, 2, 6], [0, -1, 2, -4, -11], [0, -2, 4, -1, -1], [0, -2, 4, -3, -7]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, -2, 4, -1, -1], [0, -2, 4, -3...

`Row reduce using (2),(2): `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, -2, 4, -1, -1], [0, -2, 4, -3, -7]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, 0, 0, 7, 21], [0, 0, 0, 5, 1...

`Row reduce using (1/7): `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, 0, 0, 7, 21], [0, 0, 0, 5, 15]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, 0, 0, 1, 3], [0, 0, 0, 5, 15]])

`Row reduce using (-5): `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, 0, 0, 1, 3], [0, 0, 0, 5, 15]])*` -->> `*matrix([[1, 0, -3, 2, 6], [0, 1, -2, 4, 11], [0, 0, 0, 1, 3], [0, 0, 0, 0, 0]])

`Hence, since the columns of `*A^t*` correspond to our orginal vectors, the basis is`

`comprised of `*v[1] = matrix([[1, 1, 3, 2]]), v[2] = matrix([[0, -2, -1, -2]]), v[4] = matrix([[2, 1, 2, 1]])

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, -...

A = matrix([[1, 0, 0], [0, -1/10, 0], [0, 0, 1]])