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Homework Help: Linear Transformations

  1. Jan 9, 2009 #1
    [tex]A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex]

    Let [tex]T_{A}: R^2 \rightarrow R^3[/tex] be the matrix transformation that maps a [tex]2 \times 1[/tex] column vector x in R2 into the [tex]3 \times 1[/tex] column vector Ax in R3.

    The relationship can be expressed as TA(x) = Ax


    Find a vector x in R2 whose image under TA is [tex]\left(\begin{array}{ccc}7\\0\\7\end{ar ray}\right)[/tex]



    3. The attempt at a solution

    If we write this in component form as:

    [tex]\left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex] [tex]\left[\begin{array}{ccccc} x_{1} \\ x_{2} \end{array}\right][/tex] [tex]=\left[\begin{array}{ccccc} 7 \\ 0 \\ 7 \end{array}\right][/tex]


    I could solve for [tex]\left[\begin{array}{ccccc} x_{1} \\ x_{2} \end{array}\right][/tex]

    However a linear equation in the matrix form "AX = B" can be solved easily for X if "A" is invertible and its inverse is known, you get: X = A-1B

    But [tex]A = \left[\begin{array}{ccccc} 1 & -1 \\ 2 & 5 \\ 3 & 4 \end{array}\right][/tex] is NOT square and therefore it has no inverse!

    Does anyone know how to find the solution?

    Thank you




     
  2. jcsd
  3. Jan 9, 2009 #2

    HallsofIvy

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    Science Advisor

    n equations in m unknowns with n> m MAY not have a solution at all since a linear transformation from an m dimensional vector space to another vector space maps it into an at most m dimensional subset of the second vector space. That's why there is no inverse matrix. There will be a solution only if the given vector, here <7 0 7> happens to lie in that subspace. In any case, finding an inverse matrix is seldom the best method of solving a matrix equation anyway.

    I think the simplest thing to do for as simple a system as this is just write out the equations themselve rather that use matrices:
    [itex]x_1- x_2= 7[/itex], [itex]2x_1+ 5x_2= 0[/itex], and [itex]3x_2+ 4x_2= 7[/itex]
    The first equation says that [itex]x_1=7+ x_2[/itex]. Putting that into the second equation [itex]2(7+ x_2)+ 5x_2= 14+ 7x_2= 0[/itex] so [itex]x_2= -2[/itex] and then [itex]x_1= 7- 2= 5[/itex]. Now check to see whether they satisfy the third equation: [itex]3x_1+ 4x_2= 3(5)+ 4(-2)= 15- 8= 7[/itex].

    If you must use matrices, write the augmented matrix:
    [tex]\begin{bmatrix} 1 & -1 \\ 2 & 5 \\ 3 & 4\end{bmatrix}\begin{bmatrix} 7 \\ 0 \\ 7\end{bmatrix}[/tex]
    and row reduce. Subtract 2 times the first row from the second and 3 times the first row from the third to get
    [tex]\begin{bmatrix} 1 & -1 \\ 0 & 7 \\ 0 & 7\end{bmatrix}\begin{bmatrix} 7 \\ -14 \\ -14\end{bmatrix}[/tex]
    Now divide the second row by 1, add the second row to the first row and subtract the second row from the third row to get
    [tex]\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 5 \\ -2 \\ 0\end{bmatrix}[/tex]
    The fact that the third row is now completely zeros tells us that is a correct solution.
     
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