Assignment 8
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.
Let *B* =
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* + 3*z* + 2*w* = 1
2*x* - *y* + *z* + 2*w* = 3
3*x* - *y* - *z* + 2*w* = 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* + 3*z* + 2*w* = *b*_{1}
2*x* - *y* + *z* + 2*w* = *b*_{2}
3*x* - *y* - *z* + 2*w* = *b*_{3}
Do they have a solution for every choice of *b*_{1}, *b*_{2} and *b*_{3}? If so, explain why. If not, find values of *b*_{1}, *b*_{2} and *b*_{3} so that this system does not have a solution and explain why not.
4. (2 points) Let *v*_{1} = , *v*_{2} = , *v*_{3} = and *v*_{4} = . Can every vector *u* = be written as a linear combination of *v*_{1}, *v*_{2}, *v*_{3} and *v*_{4}? Explain why or why not. Relate your answer to your answer to question 3.
5. (3 points) Find *v*_{5} = so that every vector *u* = be written as a linear combination of *v*_{1}, *v*_{2}, *v*_{3}, *v*_{4} and *v*_{5} where *v*_{1}, *v*_{2}, *v*_{3} and *v*_{4} are as in question 4. Explain why this vector *v*_{5} 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 *u*_{1}, …, *u*_{n} so that every solution of the equations in question 2 can be written as = *p* + *c*_{1}*u*_{1} + ^{} + *c*_{n}*u*_{n} for some choice of numbers *c*_{1}, …, *c*_{n}.
7. (2 points) Are the vectors in question 4 linearly independent or linearly dependent? Give reasons for your answer.
Solutions
1.
2. *B* is the augmented matrix for the equations. Using the row operation on *B* does not change the solutions. So the equations corresponding to *M* have the same solutions. The equations corresponding to *M* are *x* * * - 2*z* = 2
*y* - 5*z* + 2*w* = 1
0 = 0
The solutions to these equations are *x* = 2 + 2*z* and *y* = 1 + 5*z* - 2*w* where *w* and *z* can be any numbers. In particular, if we take *z* = *w* = 0 then we get the solution *x* = 2, *y* = 1, *z* = 0, *w* = 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* * * - 2*z* = 2
*y* - 5*z* + 2*w* = 1
0 = 1
and there will be no solution. So if *b*_{1} = 1, *b*_{2} = 3 and *b*_{3} = 6 there is no solution.
4. The vector *w* = can not be written as a linear combination of *v*_{1}, *v*_{2}, *v*_{3} and *v*_{4}. If it could then the equations *x* - *y* + 3*z* + 2*w* = 1
2*x* - *y* + *z* + 2*w* = 3
3*x* - *y* - *z* + 2*w* = 6
would have a solution. But, by the answer to question 3 it does not.
5. Let *v*_{5} have the first two components as *v*_{3} and change the third to 0, i.e *v*_{5} = . Then every vector *w* = can be written as a linear combination of *v*_{1}, *v*_{2} and *v*_{5}. This is because writing *w* as *w* = *c*_{1}*v*_{1} + *c*_{2}*v*_{2} + *c*_{5}*v*_{5} is equivalent to solving the equations *c*_{1} - *c*_{2} + 3*c*_{3} = *w*_{1}
2*c*_{1} - *c*_{2} + *c*_{3} = *w*_{2}
3*c*_{1} - *c*_{2} = *w*_{3}
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 *v*_{1}, *v*_{2} and *v*_{5}.
6. They do not have a unique solution. We saw in question 2 that the solutions are of the form *x* = 2 + 2*z*, *y* = 1 + 5*z* - 2*w* 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* .
7. They are linearly dependent. We saw in class that the solution to the system in question 2 would be unique if they were linearly independent.
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