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Math Help - Solving Systems using matrices

  1. #1
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    Solving Systems using matrices

    Hello all,

    I am currently stuck in this linear algebra problem simply because I am not sure what to do. I think I should solve by using Gauss-Jordan, but I still don't know why there are four spaces to put the solution. Here's the question:


    Solve the system:



    Well, I tried doing the Gauss-Jordan method, as mentioned before. To make it easier, I replaced X1 by x and X2 by Y. I ended up with this:

    ROW1: 1 1/2 3/2

    ROW2: 0 0 0

    I really don't know what to do or even if I am in the right path. I don't understand the way I am supposed to answer the question, wth the 4 empty boxes. I also did not found any similar problems.

    Any toughts on this? Thanks a lot in advance!
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  2. #2
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    Re: Solving Systems using matrices

    What you have discovered is you really only have "one equation" (the second equation is just the first one multiplied by -4).

    So your one equation is:

    x1 + (1/2)x2 = 3/2.

    To "solve" for this, we can only decide what x1 should be after we have picked x2 (or vice-versa, but since we usually row-reduce from "the bottom up", we designate the non-pivot rows as the "free variables").

    So let x2 = s. We then get:

    x1 + s/2 = 3/2, so:

    x1 = (3 - s)/2.

    Thus "a" solution is ((3 - s)/2,s) (which solution we get will depend on what we choose for s). We can write this as:

    (x1,x2) = (3/2,0) + s(-1/2,1).

    Let's look at what happens with "each term" in our system:

    2(3/2) + 0 = 3
    -8(3/2) - 4(0) = -12

    so (3/2,0) is a PARTICULAR solution (not the ONLY one, but one that works).

    Now the other term:

    2(-s/2) + s = -s + s = 0.
    -8(-s/2) - 4s = 4s - 4s = 0, no matter WHAT number s is.

    In other words, any multiple (by s) of (-1/2,1) is a solution to the system:

    2x1 + x2 = 0
    -8x1 - 4x2 = 0

    (called the associated HOMOGENEOUS system).

    So if we call the original matrix A, we see that:

    A((3 - s)/2,s) = A(3/2,0) + A(s(-1/2,1)) = A(3/2,0) + sA(-1/2,1) = (2(2/2) + 0,-8(3/2) - 4(0)) + s(2(-1/2) + 1, -8(-1/2) - 4) = (3,-12) + s(0,0) = (3,-12) + (0,0) = (3,-12).
    Thanks from lguto
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  3. #3
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    Re: Solving Systems using matrices

    Thanks for the reply. I still dont understand what am I being asked as an answer and where the S comes from. This is an online assignment about a subject that my professor still haven't showed in class. Do you know the name of the subject so I can find some video online to help me understand?

    I feel that I am missing something to understand your explanation, so maybe after learning it I might get it. I couldn't even find any examples similar to this online, as like I said, I don't know what that subject is called.

    Thanks for the help!
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  4. #4
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    Re: Solving Systems using matrices

    The s is called a parameter.

    Ok, say we have 3 equations in 3 unknowns. We might try elimination/substitution to get rid of one of the unknowns, and reduce to 2 equations, 2 unknowns. If that is successful, we might try to do it again, to get 1 equation in one unknown, and then use the solution for that in the 2 equations in two unknowns to solve for the second variable, and then use what we have found for the two unknowns we have solved for to finally get all 3 unknowns. This is the usual way of things.

    But sometimes, we expose a hidden relationship between the equations we didn't notice at first. Here is an example:

    x + y + z = 6
    2x - 2y + z = 1
    3x -y + 2z = 7

    Let's try to eliminate "z" by subtracting the second equation from the first, and also subtracting 2 times the first, from the second:

    -x + 3y = 5
    x - 3y = -5

    Hmmm....the two equations we get are just negatives of each other, so we don't get any new information by having two. All we know is that x = 3y - 5, we have no further information to go on. So basically we now have 1 equation in 2 unknowns. Since we have more unknowns than equations, the system of equations is what is called "under-determined", we have more than just ONE unique solution.

    Ok, so let's suppose we just pick some value for y, we'll call it t (because we don't know what it is). Now we have:

    y = t
    x = 3y - 5 = 3t - 5.

    Pick any of our original equations, and substitute back in with x = 3t - 5, and y = t. I'll use the first one:

    x + y + z = 6, which becomes:

    3t - 5 + t + z = 6.

    Solve for z, pretending we already know what "t" is (even though we don't).

    z = -3t + 5 - t + 6 = 11 - 4t

    This gives us "sort of" a solution: (x,y,z) = (3t - 5,t,11 - 4t), which we can re-write as:

    (-5,0,11) + (3t,t,-4t) = (-5,0,11) + t(3,1,-4).

    I claim the following: ANY solution of our original 3 equations is of this form.

    For example, I chose the equations so that (1,2,3) would be a solution. can we write (1,2,3) in this form? Let's see:

    (1,2,3) = (-5,0,11) + t(3,1,-4)

    (1,2,3) - (-5,0,11) = t(3,1,-4)

    (6,2,-8) = t(3,1,-4) ----> t = 2, and indeed (1,2,3) = (-5,0,11) + (6,2,-8) = (-5,0,11) + 2(3,1,-4).

    In fact, using our original 3 equations (again), we have:

    (3t - 5) + t + (11 - 4t) = 0t + 6 = 6

    2(3t - 5) - 2t + 11 - 4t = 6t - 10 - 2t + 11 - 4t = 0t + 1 = 1

    3(3t - 5) - t + 2(11 - 4t) = 9t - 15 - t + 22 - 8t = 0t + 7 = 7

    OK, now let's use matrices, and row-reduce:

    Our augmented matrix for our original system is:

    \begin{bmatrix}1&1&1&|&6\\2&-2&1&|&1\\3&-1&2&|&7 \end{bmatrix}

    Replacing row 3 with -3 times row 1 plus row 3, we get:

    \begin{bmatrix}1&1&1&|&6\\2&-2&1&|&1\\0&-4&-1&|&-11 \end{bmatrix}

    Replacing row 2 with -2 times row 1 plus row 2, we get:

    \begin{bmatrix}1&1&1&|&6\\0&-4&-1&|&-11\\0&-4&-1&|&-11 \end{bmatrix}

    Replacing row 3 with -1 times row 2 plus row 3, we get:

    \begin{bmatrix}1&1&1&|&6\\0&-4&-1&|&-11\\0&0&0&|&0 \end{bmatrix}

    This tells us we're "losing information" we really just have "at most 2" independent equations, since it reduces to a matrix with at least one zero row. Let's keep going:

    Multiplying row 2 by -1/4, we get:

    \begin{bmatrix}1&1&1&|&6\\0&1&\frac{1}{4}&|&\frac{  11}{4}\\0&0&0&|&0 \end{bmatrix}

    Replacing row 1 with row 1 minus row 2, we get:

    \begin{bmatrix}1&0&\frac{3}{4}&|&\frac{13}{4}\\0&1  &\frac{1}{4}&|&\frac{11}{4}\\0&0&0&|&0 \end{bmatrix}

    Now we are completely in reduced row-echelon form, so we can "read off the answer".

    The last row tells us z can be anything: since 0z = 0 is true for any number z. To avoid confusion with what we did above, let's write z = s.

    The second row tells us y + z/4 = 11/4, or:

    y + s/4 = 11/4, or:

    y = (11 - s)/4.

    The first row tells us:

    x + 3z/4 = 13/4, or:

    x = (13 - 3s)/4.

    so (x,y,z) = ((13 - 3s)/4, (11 - s)/4, s) = (13/4,11/4,0) + s(-3/4,-1/4,1).

    Now, this certainly LOOKS different than what we did above, right?

    But, suppose that having chosen s, we chose t = (11 - s)/4. Our original solution was (see above):

    (x,y,z) = (-5,0,11) + t(3,1,-4).

    Using t = (11 - s)/4, this becomes:

    (x,y,z) = (-5,0,11) + [(11 - s)/4](3,1,-4)

    = (-5,0,11) + (11/4)(3,1,-4) + (s/4)(-3,-1,4)

    = (-5,0,11) + (33/4,11/4,-11) + s(-3/4,-1/4,1)

    = (13/4,11/4,0) + s(-3/4,-1/4,1).

    On the other hand, if after having chosen t, we choose s = 11 - 4t, we get:

    (13/4,11/4,0) + (11 - 4t)(-3/4,-1/4,1) =

    (13/4,11/4,0) + (-33/4,-11/4,11) - 4t(-3/4,-1/4,1) =

    (-20/4,0,11) - t(-3,-1,4) = (-5,0,11) + t(3,1,-4), showing that we get the same set of solutions either way.

    ******

    In general, when solving a matrix equation Ax = b, answers are of the form:

    x = particular solution + homogeneous solution.

    A homogeneous solution is a solution to:

    Ax = 0.

    If x0 is some vector for which:

    Ax0 = b, and u is ANY solution to Ax = 0, then for x = x0 + u

    A(x) = A(x0 + u) = Ax0 + Au = b + 0 = b.

    In the example I just gave, the "parameter part" (the vector multiplied by s, or t) represents the homogeneous part: for example, (3,1,-4) is a solution to:

    x + y + z = 0
    2x - 2y + z = 0
    3x - y + 2z = 0

    As you can verify yourself, (-3/4,-1/4,1) is also a homogeneous solution (not surprising, since it is -1/4 times the other homogeneous solution).

    Both (5,0,11), and (13/4,11/4,0) are "particular solutions" (ones for which A(x,y,z) = (6,1,7)).

    *********

    Finally, a *very abstract* look at what is going on:

    In some sense, the rank of a matrix tells you "how big" the range of A is. In this case, rank(A) = 2, which tells us our "target space" (the values A(x,y,z) can take on) has dimension 2, it forms a plane in 3-space.

    This plane is determined by the two lines going from the origin through (1,2,3) and (1,-2,-1), respectively. Fortunately for us, the point (6,1,7) lies in this plane:

    (6,1,7) = (11/4)(1,2,3) + (13/4)(1,-2,-1). If this were not so, we could not solve the system of equations at all!

    The "solution space" in this case is a line, it passes through the points (-5,0,11) and (13/4,11/4,0), this line is mapped by A to the single point (6,1,7), since A maps any scalar multiple of (3,1,-4) to (0,0,0).

    The line L = (-5,0,11) + t(3,1,-4) = (13/4,11/4,0) + s(-3/4,-1/4,1) is called a COSET of the line through the origin L0 = t(3,1,-4). Cosets of L0 are just parallel lines (like a stack of straws all pointing the same direction).

    The vector (-5,0,11) or (13/4,11/4,0) is called the translation vector of L, it tells us where to move the (homogeneous) line L0, so that we map to (6,1,7) instead of (0,0,0).

    One thing that is NOT obvious, is that any two point on L differ by a point on L0, for example:

    (-5,0,11) - (13/4,11/4,0) = (-33/4,-11/4,11) which is just the vector (3,1,-4) multiplied by the scalar -11/4.

    L0 is also called the null space, or kernel of A, it represents all the points in 3-space that A "shrinks to 0" or annihilates.

    *********

    Finally, the part of mathematics where this is studied is: the theory of linear equations, or more commonly, linear algebra. The objects that we "do stuff to" are called VECTORS, and they live in VECTOR SPACES. The functions from one vector space to another that preserve "vector operations" are called LINEAR TRANSFORMATIONS, and are, for the most part, equivalent to matrices. I do not think I am understating the case to say that linear transformations are one of the most important mathematical concepts to have been brought to light in the last 300 years. That are THAT useful, in more areas than I can even count.

    I feel you might find some material that might match your current level of understanding at:

    https://www.khanacademy.org/

    There are also several good Khan Academy videos on youtube, like this one, for example:
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