(BLOCK OF 1's & 0's WITH| JJJJJJ UU UU NN N K K | Linear stuff ) (THE 1's MOVING DOWN AND| JJ UU UU N N N KK | from original) (TO THE RIGHT. | JJJJ UUUUUU N NN K K | right sides ) (--------------------------------------------------------------------) ( MAYBE HERE SEVERAL | More such ) ( ROWS OF 0'S | linear stuff )Please observe that the lower right-hand corner now plays the part of the compatibility conditions which must be satisfied. All of those linear "fragments" must be equal to 0 if the original system has solutions. Now if these are 0, we can "read off" solutions in much the same manner as the example. The block labeled JUNK in fact tells us with its width how many free parameters there are in the solutions. Notice that the JUNK block could have size 0 (for example, consider the silly system x1=567,672, already in RREF!) and in that case the system would have only one solution.
Problem 3
Are the vectors (4,3,2) and (3,2,3) and (-4,4,3) and (5,2,1) in
R3 linearly independent? Now I wrote the vector
equation:
A(4,3,2)+B(3,2,3)+C(-4,4,3)+D(5,2,1)=0 (this is (0,0,0) for this
instantiation of "linear independence") which gives me the system:
4A+3B-4C+5D=0 (from the first components of the vectors)
3A+2B+4C+2D=0 (from the second components of the vectors)
2A+3B+3C+1D=0 (from the third components of the vectors)
and therefore would need to row reduce
( 4 3 -4 5 ) ( 3 2 4 2 ) ( 2 3 3 1 )I started to do this, but then ... thought a bit. My goal was to get the RREF of this matrix, and use that to argue about whether the original system had solutions other than the trivial solution.
What can these RREF's look like? Let me write all of the RREF's possible whose first column (as here) has some entry which is not zero.
( 1 0 0 * ) ( 1 0 * 0 ) ( 1 0 * * ) ( 1 * 0 0 ) ( 1 * 0 * ) ( 1 * * 0 ) ( 1 * * * ) ( 0 1 0 * ) ( 0 1 * 0 ) ( 0 1 * * ) ( 0 0 1 0 ) ( 0 0 1 * ) ( 0 0 0 1 ) ( 0 0 0 0 ) ( 0 0 1 * ) ( 0 0 0 1 ) ( 0 0 0 0 ) ( 0 0 0 1 ) ( 0 0 0 0 ) ( 0 0 0 0 ) ( 0 0 0 0 )In all of these 3 by 4 matrices, the entry * stands for something which could be any number, zero or non-zero. I hope I haven't missed one! In every one of the matrices above, there will be non-trivial solutions to the related homogeneous system. For example, in the first matrix, D is free to be any value, and the equations can be made correct by selecting the correct values of A and B and C. In the second matrix, D=0 certainly (because of the last row) but C can be any value, and A and B can be selected to make the other two equations correct. In each row where there are *'s, those variables are free, and the leading 1 in the row represents a variable whose value can be chosen to make the corresponding equation correct. Therefore all of these RREF's represent homogeneous systems with non-trivial solutions. And notice that if the first column had been all 0's then A could have any value. So, in fact, a homogeneous system having 4 variables and 3 equations will always have a non-trivial solution.
| A homogeneous system with more variables than equations always has an infinite number of non-trivial solutions. |
Many problems in engineering and science turn out to be successfully modeled by systems of linear equations. My examples so far have mostly been"inspired" by the things I believe you will see in studying ODE's and PDE's, and have also been influenced by some of the primary objects in numerical analysis. The textbook has a wonderful diagram, a sort of flow chart, for solving linear systems on p. 366. I think our discussions in class have been sufficient for you to verify the diagram, and the diagram contains just about everything you needs to know about the theory behind solution of systems of linear equations. The only remaining definition you need is that of the rank of a matrix. The rank is the number of non-zero rows in the RREF of the matrix.
Here is my version of the diagram in HTML:
For m linear equations in n unknowns AX=B
Two cases: B=0 and B not 0. Let rank(A)=r.
AX=0
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v v
Unique sol'n: X=0 Infinite number of sol'ns.
rank(A)=n rank(A)<n, n-r
arbitrary parameters
in the sol'n
AX=0, B not 0
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v v
Consistent Inconsistent
rank(A)=rank(A|B) rank(A)<rank(A|B)
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v v
Unique solution Infinite number of sol'ns
rank(A)=n rank(A)<n-r
arbitrary parameters
in the sol'n
Let try to offer a gloss on this diagram, which maybe looks simple but
covers so many different situations which you will enounter.
gloss 1. [Linguistics][Semantics] a. an explanatory word or phrase inserted between the lines or in the margin of a text.First, consistent means the system has solutions, and inconsistent means there are none. Here is a tabular listing of the alternatives, if you find this more palatable. The B=0 case, the homogeneous system, always has the trivial solution (used for, say, deciding linear independence). So the B=0 case is always consistent. Two alternatives can occur:
| AX=0: m equations in n unknowns; B=0 | |
|---|---|
| I | II |
| Unique sol'n: X=0 rank(A)=n |
Infinite number of solutions. rank(A)<n, n-r arbitrary parameters in the solutions |
When B is not zero then we've got:
| AX=0: m equations in n unknowns; B not 0 | ||
|---|---|---|
| Consistent rank(A)=rank(A|B) | Inconsistent rank(A)<rank(A|B) | |
| III | IV | V |
| Unique solution rank(A)=n |
Infinite number of solutions
rank(A)<n-r arbitrary parameters in the solutions |
No solutions |
By the way, I do not recommend that you memorize this information. No one I know has done this, not even the most compulsive. But everyone I know who uses linear algebra has this installed in their brains. As I mentioned in class, I thought that a nice homework assignment would be for students to find examples of each of these (in fact, there have already been examples of each of these in the lectures!). The problem with any examples done "by hand" is that they may not reflect reality. To me, reality might begin with 20 equations in 30 unknowns, or maybe 2,000 equations ....
So I just gave 6 examples which we analyzed in class. These examples are (I hope!) simple (m=number of equations; n=# of variables; r=rank(A), where A is the coefficient matrix):
( 2 3)~( 1 3/2)~( 1 0 ) (-5 7) ( 0 29/2) ( 0 1 )I think you could already have seen that the rows of this 2-by=2 matrix were linearly independent just by looking at it (don't try "just looking at" a random 500-by-300 matrix!), so r=2. This is case II. There is a unique solution, x=0 and y=0, the trivial solution.
( 2 3 | 1)~( 1 3/2 | 1/2)~( 1 0 | 1/2 - (3/2)·(9/29)) (-5 7 | 2) ( 0 29/2 | 9/2) ( 0 1 | 9/29 )we are in case III, with exactly one solution which row reduction has produced: x=1/2-(3/2)·(9/29)) and y=9/29.
( 2 3 | 1) ( 1 3/2 | 1/2) ( 1 0 | 1/2 - (3/2)·(9/29)) (-5 7 | 2)~( 0 29/2 | 9/2)~( 0 1 | 9/29 ) ( 4 5 | 3) ( 0 -1 | 1 ) ( 0 0 | 1-(9/29) )I am lazy and I know that 1-(9/29) is not 0, so the row reduction showed that rank(A)=2<rank(A|B)=3: case V, with no solutions.
Heuristic, the word
adj. 1. allowing or assisting to discover. 2. [Computing] proceeding to a solution by trial and error. "heuristic method"[Education] a system of education under which pupils are trained to find out things for themselves.heuristic gives 882,0000 hits on Google and adding engineering to that reduces the number to "only" about 181,000. Indeed.