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Linear Algebra

  1. Jan 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Choose h and k such that the system has
    a) no solution
    b) a unique solution and,
    c) many solutions.

    Give separate answers for each part.

    x1 + hx2=2
    4x1+8x2=k


    2. Relevant equations



    3. The attempt at a solution
    I set up the matrix

    [1 h 2 ]
    [4 8 k ]

    and I multiplied the top row by -4 and added the 3rd row to it to get

    [ 1 ----- h ----- 2 ---]
    [ 0 (-4h+8) (-8+k) ]

    To get part A) -no solution-, I figured that the 2nd column cannot equal 0, therefore if h=2, and k does not equal 8, then there is no solution.

    Here is where I get lost...
    I didn't get these answers that the book says:

    For part B), unique solution, h does not equal 2....but what does that mean? If h doesn't equal 2 then that means it can take on any value besides 2, therefore k can be almost any value as well.

    For part C), many solutions, that books says h=2 and k=8
    So that means for the 2nd row you would get 0=0....so wouldn't we just disregard that? then we are left with the 1st row saying 1(x1)+2(x2)=2

    I don't understand parts B and C.

    Another question: Am I right to assume that we're just looking at -4h+8 here? Or should I also be considering -8+k??

    Thanks!
     
  2. jcsd
  3. Jan 24, 2008 #2
    Do you know about determinants and Cramer's rule?
     
  4. Jan 24, 2008 #3
    Nope I do not.
     
  5. Jan 24, 2008 #4
    Ok! Solve the first equation for [tex]x_1[/tex] and substitude the result into the second one.
    In order to solve the reduced equation for [itex]x_2[/itex] it's coefficient must be non-zero.
     
  6. Jan 24, 2008 #5

    Defennder

    User Avatar
    Homework Helper

    If you have learned what is reduced-row echelon forms for a matrix, then you would observe that in order for the unknowns to be solvable for 1 solution, then the reduced-row echelon form of the matrix is the identity matrix. You do know how to use elementary row operations, do you?

    For the part C, note that is required that you obtain a row of 0s in your matrix for its reduced-row echelon form.
     
  7. Jan 25, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That means exactly the opposite of what you said in part A. "If h= 2 and k does not equal 8, then there is no solution" If h is any number other than 2, then you can solve for x2 and so for x1. You will have exactly one solution.

    WHY did you say in part A "and k does not equal 8, then there is no solution"? In part A, if h= 2 and k is not 8 then you have, for the last row, 0 0 8-k, which is equivalent to the equation 0x1+ 0x2= 8-k. If 8-k is not 0, there are NO values of x1 and x2 which will satisfy that. If 8-k= 0, you have 0 0 0, equivalent to 0x1+0x2= 0 which is true for ALL x1, x2. You are, then left with the 1st row as you say. You could pick any number for, say, x1 and solve for x2: there are an infinite number of solutions.

    To answer what question? As long as -4h+ 8 is not 0, then there exist a unique solution no matter what -8+ k is. But is -4h+ 8= 0, then -8+ k becomes important. If, in that case, -8+ k= 0, there are an infinite number of solutions. If -8+ k is not 0, then there exist NO solution.
     
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