Due: Monday, March 14. Nothing accepted after Tuesday, March 15. 10% off for being late

Please work by yourself on this assignment and all the other assignments that you turn in as part of your grade. Please see me if you need help.

1. (2 points) Show that B can be transformed by elementary row operations to the reduced row echelon form matrix M = .

2. (2 points) Consider the system of equations

x - y + 3z + 2w = 1

2x - y + z + 2w = 3

3x - y - z + 2w = 5

Do they have a solution? If so, find one. If not, explain why not.

3. (2 points) Consider the system of equations

x - y + 3z + 2w = b₁

2x - y + z + 2w = b₂

3x - y - z + 2w = b₃

Do they have a solution for every choice of b₁, b₂ and b₃? If so, explain why. If not, find values of b₁, b₂ and b₃ so that this system does not have a solution and explain why not.

4. (2 points) Let v₁ = , v₂ = , v₃ = and v₄ = . Can every vector u = be written as a linear combination of v₁, v₂, v₃ and v₄? Explain why or why not. Relate your answer to your answer to question 3.

5. (3 points) Find v₅ = so that every vector u = be written as a linear combination of v₁, v₂, v₃, v₄ and v₅ where v₁, v₂, v₃ and v₄ are as in question 4. Explain why this vector v₅ works.

6. (2 points) Do the system of equations in question 2 have a unique solution. If so, explain why. If not, find vectors p and u₁, …, u_n so that every solution of the equations in question 2 can be written as = p + c₁u₁ + ^ + c_nu_n for some choice of numbers c₁, …, c_n.

7. (2 points) Are the vectors in question 4 linearly independent or linearly dependent? Give reasons for your answer.

Solutions

1.

0 = 0

3. In question 2, the last equation in the equations corresponding to the matrix M was 0 = 0 which was true no matter what the values of x, y, z and w are. If instead this last equation was 0 = c where c is non-zero, then there would be no solution. So if we change the 5 on the right hand side of equation 3 in the original set of equations then the augmented matrix will be B' = . After doing the same row operations as in the solution to question 1, the matrix B' will be transformed to M' = . The corresponding equations will be x - 2z = 2

0 = 1

4. The vector w = can not be written as a linear combination of v₁, v₂, v₃ and v₄. If it could then the equations x - y + 3z + 2w = 1

2x - y + z + 2w = 3

3x - y - z + 2w = 6

would have a solution. But, by the answer to question 3 it does not.

5. Let v₅ have the first two components as v₃ and change the third to 0, i.e v₅ = . Then every vector w = can be written as a linear combination of v₁, v₂ and v₅. This is because writing w as w = c₁v₁ + c₂v₂ + c₅v₅ is equivalent to solving the equations c₁ - c₂ + 3c₃ = w₁

2c₁ - c₂ + c₃ = w₂

3c₁ - c₂ = w₃

According to the discussion in class, this will always have a solution if when we reduce the coefficient matrix to reduced row echelon form, there is no row that is all zero. The coefficient matrix is A = . One has . It is in reduced row echelon form so there always is a solution so every vector can be written as a linear combination of v₁, v₂ and v₅.

6. They do not have a unique solution. We saw in question 2 that the solutions are of the form x = 2 + 2z, y = 1 + 5z - 2w where w and z can be any numbers. If we take z = w = 0 then we get the solution x = 2, y = 1, z = 0, w = 0. If we take z =1, w = 0 then we get the solution x = 4, y = 6, z = 1, w = 0. In general = + z + w .