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3x3 matrix a1+2a2-a3=0

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data
    If A is a 3 x 3 matrix a1+2a2-a3=0, then A must be singular.
    I have the answering being true but how do I prove it?
  2. jcsd
  3. Feb 7, 2010 #2
    Assuming a_i are columns or rows, then sure. Let's assume you row/column reduce something as far as you can.

    We have two cases:
    Case 1:
    You end up with the identity. Taking the determinant, you'll get a nonzero number which will be affected by the row/column operations you did (like if you swapped rows and whatnot).

    Case 2:
    You have a row/col of zeroes somewhere.
    Taking the determinant of this will give you zero (hence singular).

    Since a_1 + 2a_2 - a_3 = 0, this implies that a_1 + 2a_2 = a_3.

    What case is the matrix in? Why?
  4. Feb 7, 2010 #3
    We can't prove it using determinants. The equation is in the form Ax=b. Where x is the column vector 1,2,-1.
  5. Feb 7, 2010 #4
    What does that have to do with anything? According your original post, this deals only with the matrix A. What does that particular equation have anything to do with whether A is singular or not?
  6. Feb 7, 2010 #5
    Because that column vector is used in proving the singularity but I don't know how to do it.
  7. Feb 7, 2010 #6
    I don't understand the question. The vector in question has nothing to do with that.

    What are the ai? If they are columns/rows, you don't need that column vector. What is b, anyways?
  8. Feb 7, 2010 #7
    Column vector b is the 0 vector. It has to do with homogeneous equations have trivial and solutions. That is how A is suppose to be proving for the question.
  9. Feb 7, 2010 #8
    Oooooh! I got you. This seems simple enough.

    Consider any general system Ax = b.

    If A was invertible, what does this tell you about x?
  10. Feb 7, 2010 #9
    x is the inverse if and only if b is the I
  11. Feb 7, 2010 #10
    No no. x can't be the inverse of anything because x is a column vector. x is a solution to this equation. What kind of solution, though? Hint: Is it the only solution?
  12. Feb 7, 2010 #11
    Nontrivial solution
  13. Feb 7, 2010 #12
    There are no "trivial" solutions if b is not the zero vector. So, if A is invertible and b is not the zero vector, does there exist a y such that Ay = b and x =/= y ? That is to say, is x unique?
  14. Feb 7, 2010 #13
    It is giving that b is the 0 vector. I am not sure if it is unique or how to show if it isn't.
  15. Feb 7, 2010 #14
    The fact b is the zero vector is the key part in the proof, because it guarantees you a solution. It guarantees you the trivial solution of x = 0. According to the problem at hand, you also have another solution of the form x = <1,-2,1>. Therefore, x is not unique.

    Is there any way of connection uniqueness of solutions with whether A is invertible?
  16. Feb 7, 2010 #15
    The system Ax=b of n linear equations in n unknowns has a unique solution if and only if A is nonsingular. Since x can be the 0 vector and vector <1,2,-1>, the solution isn't unique; therefore, A must be singular.
  17. Feb 7, 2010 #16
  18. Feb 7, 2010 #17
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