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November 16th, 2012, 08:49 AM   #1
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matrix simplification

my observation-

while solving simultaneous equations by matrix method, or simplifying a matrix,
a row can be multiplied by a no. and can be added or subtracted from another row for simplification. now, in a matrix, there is no way a single row can be multiplied by a constant. if a matrix is multiplied by a no., all its elements are multiplied by that no.
on what basis a row is multiplied by a no. in a matrix? how does the whole matrix gets affected by this ( matrix has no value, determinant has)?


i would be thankful for any help.
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November 16th, 2012, 12:30 PM   #2
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Re: matrix simplification

Quote:
Originally Posted by philosopher
my observation-

while solving simultaneous equations by matrix method, or simplifying a matrix,
a row can be multiplied by a no. and can be added or subtracted from another row for simplification. now, in a matrix, there is no way a single row can be multiplied by a constant. if a matrix is multiplied by a no., all its elements are multiplied by that no.
on what basis a row is multiplied by a no. in a matrix? how does the whole matrix gets affected by this ( matrix has no value, determinant has)?


i would be thankful for any help.
I may not be completely understanding your question, so I apologize if my answer isn't helpful but I'll do my best. It sounds like what you're asking is how is it possible to preserve information about a system of equations in a matrix by multiplying a single row by a scalar rather than the entire matrix (in which case every entry would be adjusted by that scalar).

One way to think about this is to consider what exactly is going on in a system of equations / matrix. Let's say we have a 3 x 4 augemnted matrix with only real entries. The first row of that matrix can be viewed as the following linear equation (potentially, I just chose random numbers here): . Now what happens if you multiply that equation by a scalar, let's say, 2? Then we get the equation which is "equivalent" to our original row because we've multiplied both sides of the equation by the same scalar quantity.

Thinking about this in terms of a matrix: A matrix is simply a way to visualize a linear system of equations. If you multiply one row of a matrix (one equation of a linear system) by a scalar and then add that row (equivalent linear equation) to another row (linear equation) in the matrix then you're still preserving the numerical "relationship" between the variables of the linear system. The proof for this is somewhat difficult, but you shouldn't need to grow through that quite yet -- it should suffice for you to understand the basic mechanics and logic of what is going on. You can try writing out your own linear system and verifying that this is true.

Remember, a "matrix" is just a way to organize and visualize the coefficients of a linear system. While adding a multiple of one row to another does change the relationship between the entries in those rows, it doesn't change the relationship between the variables that you are trying to solve for in the linear system which is why it works. I hope this helps.
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November 17th, 2012, 07:44 AM   #3
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Re: matrix simplification

thankyou decave, that was certainly helpful and i must say that your explanation is excellent.
that explains if the matrix is developed from a set of simultaneous eqns., now what if we have developed our matrix from a set of vectors?
moreover, what i think now is the matrix like thing that is developed from simultaneous eqns. is not actually matrix, it is a set of numbers looking like one on which we perform simplification as its usually done on a set of simultaneous equations itself.
but again, there are questions like simplification of matrix that take up a similar course of row simplification.
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November 17th, 2012, 08:59 AM   #4
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Re: matrix simplification

Thank you for the kind words. Exactly -- that's what the basic premise of a matrix is -- a system of linear equations (hence why the course is called linear algebra). Of course, we also use matrices for other things like representing linear transformations, solving dynamical systems, etc.

However, when the columns of a matrix is composed of vectors than viewing the matrix as a system of linear equations can still be very helpful. For instance, let's say you have an n x n matrix A that is composed of the following vectors in . Then row reducing A under the homogeneous equation (finding the values of when A = 0) can tell us if the vectors are linearly independent.

Your core intuition, however, is correct. To ask, "Why can we multiply a row of a matrix by a scalar? Doesn't this change the 'value' (whatever that means) of the matrix?" Firstly, you're correct, but you have to realize that much of the mathematics that is done with matrices is done "by definition". In other words, we, as mathematicians, specify exactly how and why we do certain operations with matrices. Multiplying a row by a scalar covers up some information (it changes the columns of the matrix that are a basis of a vector space) and we take that into consideration when doing row operations. However, doing row operations can also give us a lot of information as well. For instance, it can tell us when a set of vectors are linearly independent (which in and of itself gives us a WHOLE lot of other useful information as well).

In summation, when doing matrix operations, don't think of it in terms of, "Why is this allowed because doesn't it change the value of the matrix?" Rather, think of it in terms of, "What aspect of matrices and linear systems am I affecting by doing this operation?" Hope that helps, my friend! Let me know if you need anything else clarified.
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November 18th, 2012, 07:32 AM   #5
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Re: matrix simplification

Many thanks Decave.
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November 18th, 2012, 03:09 PM   #6
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Re: matrix simplification

If one row of a matrix is multiplied by a number, you do NOT have the number times that same matrix. That doesn't mean that you cannot multiply a single row of a matrix by a number! You get a different matrix but the linear equations you got the matrix from still have the same solutions, just as when you multiply both sides of an equation by a number, you get a different equation, but one that still has the same solution.
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November 19th, 2012, 05:14 PM   #7
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Re: matrix simplification

thankyou HallsofIvy.

from this whole discussion, i concluded that matrix is not that a rigid concept. its a way to manipulate systems easily (of course, there are set of rules for that).
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