Does the matrix
0 1 0 -15
0 0 1 7
has a unique solution?
Thank you in advance
if what you really mean is:
"does
$\displaystyle \begin{bmatrix}0&1&0\\0&0&1 \end{bmatrix} \begin{bmatrix}x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix}-15\\7 \end{bmatrix}$
have a unique solution?"
the answer is: no.
(1,-15,7) is a solution, and so is (2,-15,7), and these are different. there are LOTS of different ways to look at this:
1. we have 3 unknowns, and 2 equations. so IF the equations are consistent (and they are), the system is "under-determined" (more unknowns than equations).
2. the rank of the matrix is 2, which is less than the dimension of the domain.
3. as a rule: general solution = any specific solution + ALL homogeneous solutions. the homogenous case is found by replacing (-15,7) with (0,0) and solving THAT case.
this is equivalent to finding the null space of the matrix. it is clear in this case, that the null space is all vectors of the form (t,0,0) (where t can be any real number, assuming that is the field you are working in).
"free variables" correspond to the COLUMNS without a pivot (leading 1). in this case, that is column 1, so "x_{1} is free".
let me explain number 2. above a bit more: when a matrix A "acts" on a vector space R^{n}, by taking x to Ax, it takes an n-vector to an m-vector (supposing that A is mxn). so A(R^{n}) (the image space of A) is some subspace of R^{m}. now if m < n (more columns than rows), A takes R^{n} to a "smaller" space.
you can't take a "bigger" thing to a "smaller" thing without some "collapsing" occuring (two different vectors in R^{n} have to wind up "with the same image", since R^{m} doesn't have enough dimensions to reproduce all of R^{n} faithfully). this is what causes "non-uniqueness" (sorry for all the quotes, but i'm not speaking precisely, here).
what rank(A) tells us, is "how big the image space is". in this case, it has dimension 2: and a basis for im(A) is: {(1,0),(0,1)} (in other words A is of FULL rank, the image of A is all of the target space...A is surjective, or "onto" R^{2}. note that these are precisely the linearly independent COLUMNS of A: the image of A is also known as the COLUMN space).
and THAT means that Ax = b is going to have a solution (in fact, many solutions) for any b, including b = (-15,7).
going back to number 3: since a homogeneous solution is: (t,0,0) for any (real) number t, and (0,-15,7) is a PARTICULAR solution,
the general solution is: (t,0,0) + (0,-15,7) = (t,-15,7), where we are "free" to pick t as we please.
I'm grateful to you for your profound explanation...!!!
If you let me just ask one more question:
according to
if A wasin other words A is of FULL rank, the image of A is all of the target space...A is surjective, or "onto" R2.
010
000
and b was (-15,0)
I should say "the image of A is just R"?
no, because the image of A ISN'T R, it's a 1-dimensional subspace of R^{2}.
these are isomorphic (they "act alike") as VECTOR SPACES, but not as "other things".
for example, in R, we have multiplication, and "fractions". what is:
1/(a,0) ? what does that even MEAN (how do you "divide by a vector")?