Can Linear Algebra Solve for the Kernel in a System of Equations?

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in the link i wrote the question and the way i tried to solve it

but i got a really weird answer in the first part

and in the second part i solved it too
but i am not sure if my theoretical knowledge about finding ker and I am are righthttp://img512.imageshack.us/my.php?image=img8126fw0.jpg
 
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#1: The dimension of the nullspace of the matrix (which is the dimension of the kernel of the transformation that the matrix represents) is be the number of free variables. So 3a+b-c must equal zero.

#2: Remember that the dim ker(T) + dim Im(T) = dim U, where dim U = dim R^4 = 4. Also keep in mind that dim Im(T) is the rank of the matrix that represents T.

When the dimension of the kernel is minimum, the dimension of the image (the rank of A) is maximum and vice versa.
 
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i know all of this
but in normal matrices that's how i would find the ker(t)
what the problem with that??

and about the second part did i make it right??
 
You find the kernel of T by solving the equation Tx= 0. In terms of a matrix with "unknowns" such as a, b, c, that means row reducing the matrix as much as you can, then choosing a, b, c so as to make rows all 0. The "maximal kernel" is, of course, the kernel of highest dimension. You find the maximal kernel by choosing a, b, c to make as many rows as possible all 0's. The second part of the question asks you to find a, b, c so as to make the image "maximal"- i.e. as high a dimension as possible. Since nullity+ rank= dimension of target space, (nullity= dimension of kernel, rank= dimension of image), you make the range maximal by making the kernel minimal: choose a, b, c so as to make as few rows as possible all 0.
 
that what i did
but i get an expresion of
t(3a-b+c)=0

in order to the line to be all zeros
i need to say
"3a-b+c=0"

the problem with that solution is that in order to find the ker(T)
we need to SOLVE those equations
why arent we solving them??

about the second part i don't want to use the law of dimention
i solved it in a certain way
is that ok what i did??
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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