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Linear Algebra T/F.

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data

    I want to make sure I understand this.
    True False.
    1. Every matrix is row equivalent to a unique matrix in echelon form.
    2. If u,v, and w are nonzero vectors in R2 then w is a linear combination of u and v.
    3. If a system of linear equations has no free variables, then it has a unique solution.
    4. If a system of linear equations has two different solutions, it must have infinitely many solutions.
    5. If matrices A and B are row equivalent, they have the same reduced echelon form.
    6. If an n x n matrix A has n pivot positions, then the reduced echelon form of A is the n x n identity matrix.
    7. A linear transformation is a function.
    8. If a system Ax = b has more than one solution then so does the system Ax = 0.
    Question 8 answers
    9.The equation Ax = 0 has the trivial solution if and only if there are no free variables.
    10. If {u, v, w} is linearly independent, then u, v, and w are not in R2.
    11.If BC = BD then C = D. For matrix
    12. If AB = I, then A is invertible
    13. If AB = BA and A is invertible, then A^-1*B = B*A^-1.
    14.If b = [1 0 0]^T (i.e b is a column vector) and if A is a 3 x 3 matrix such that Ax = b has a unique solution, then A is invertible.
    15.If AB = C and C has 2 columns, then A has 2 columns.

    3. The attempt at a solution
    1. True.
    2. False. This is not true all the time.
    3. False. It could be inconsistent with a pivot in last column.
    4. True. Linear dependence?
    5. True. Because two matrices are equivalent if they have the same solution set.
    6. False. Not too sure on this one. Isn't an identity matrix formed by AC=CA=I(identity matrix)
    7. True. Ax=b=T(x)
    9. True. Do they mean a trivial solution or solutions.?
    10. False. At least one of them could be in R^2.
    11. False. Have a feeling because multiplication properties of matrices are not the same the other way around.
    12. True. Because A is invertable if and only if A is row equivalent to I(identity matrix).
    13. True. Can be reduced to identity matrix.
    14. True. Got a hunch that it is true, but not completely sure that it is true.
    15. True. Multiplication rules.

    Please tell me my wrong doings and tell me why please. Thanks.
  2. jcsd
  3. Feb 13, 2008 #2
  4. Feb 14, 2008 #3
  5. Feb 14, 2008 #4


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    Impatient, aren't you?

  6. Feb 14, 2008 #5
    2. False. I am pretty sure because one of the vectors could not be a linear combination of the two.

    4. I am actually having doubts on.

    6. False. I still think because an identity matrix can't be formed all the time.

    How do the others look?

    12. This should be false actually because we do not know if BA=I
  7. Feb 14, 2008 #6


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    Science Advisor
    Homework Helper

    2. What about u = 2v ?

    4. Write the system as Ax = b. When are there no solutions, when is there exactly one, and when is there more than one? If there is more than one, must there necessarily be infinitely many?

    6. Consider this 3x3 matrix
    [tex]\begin{pmatrix} 1 & 3 & 2 \\ 0 & 4 & 6 \\ 0 & 0 & 1 \end{pmatrix}[/tex]
    First use row two to get zero in the first row, second column
    [tex]\begin{pmatrix} 1 & 0 & * \\ 0 & 4 & 6 \\ 0 & 0 & 1 \end{pmatrix}[/tex]
    Now use row three to get zero in the third column
    [tex]\begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}[/tex]
    and divide the second row by four.
    Can you generalize this algorithm to any dimension, or point out where it would go wrong?

    12. Try sandwiching AB between B on the left and B^{-1} on the right.
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