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In linear algebra we know that if $A$ (is an invertible matrix) is given, then the solution for the system $Ax=b$ is $x=A^{-1}b$ for any $b$. Can any one tell me what will be the solution $x$ (if it exist)if $A$is non invertible matrix for any $b$ in the range of $A$? In other words, can we write $x$ as a function of $b$?

Can any one help me in this issue?

Thanks to everyone.

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If $Ax=b$ and $A$ is not invertible then there are two possibilities.

(1) There is no solution.

(2) There are an infinite number of solutions.

Since you are dealing with case (2), you have to use techniques @Ataraxia mentioned.

For example consider \begin{equation} x+2y=4\\ 2x+4y=14 \end{equation} We can represent this in matrix format as follows: $$ \begin{pmatrix} 1&2\\ 2&4\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}=\begin{pmatrix}7\\14\\\end{pmatrix} $$

Well, $det(A)=0$ hence it is invertible. In fact the two equations are the same. Hence the solutions can be described using the parameter $t$ as follows: $$ x=t\\ y=\frac{1}{2}(7-t) $$

This describes an infinite number of solutions since $t\in\mathcal{R}$. It describes a line. But as already mentioned, you can't write an explicit solution for $x$ in terms of $b$.

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  • $\begingroup$ what is these techniques? what does techniques @Ataraxia mentioned means? $\endgroup$ – LoveMath Aug 1 '13 at 5:36
  • $\begingroup$ I am referring to the answer people and the techniques of Gaussian or Gauss Jordan elimination. $\endgroup$ – felasfa Aug 1 '13 at 6:15
  • $\begingroup$ I know how to solve this system if b is just any vector with numbers but what I don't know and want to know exactly is how to find x in terms of $b=<a,b,c,..>$ $\endgroup$ – LoveMath Aug 1 '13 at 6:38
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When $A$ is not invertible, there is not one single solution, but an entire family of solutions. This is why the inverse does not exist, because $A\vec{x}$ is not an injective mapping in this case. Other techniques must be used, such as Gaussian or Gauss Jordan elimination.

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  • $\begingroup$ yes, there is.But what I want to know how to find at least one $x_{0}$ of these solutions that satisfy $Ax_{0}=b$? what is the way to find that, and if we can write $x$ as a function of $b$ $\endgroup$ – LoveMath Aug 1 '13 at 5:33
  • $\begingroup$ @LoveMath With Gaussian elimination or Gauss Jordan elimination. Or is there something else you're looking for? $\endgroup$ – Ataraxia Aug 1 '13 at 5:35
  • $\begingroup$ well, I want to get this solution symbolically, that is the entire solution or as I said how can we write x as a function of b for example? $\endgroup$ – LoveMath Aug 1 '13 at 5:41
  • $\begingroup$ @LoveMath Well the problem with that is, by definition can't have a it as a function of anything, because a function must have one output for one input. For a non-invertible matrix, any such relation wouldn't be one to one, because there would be an infinite number of outputs for one input. $\endgroup$ – Ataraxia Aug 1 '13 at 5:59
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    $\begingroup$ @LoveMath With Gaussian elimination...see the link. Sorry if I sound like a broken record, but I'm just not sure why you don't find elimination to be a sufficient method =/ $\endgroup$ – Ataraxia Aug 1 '13 at 6:28
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When $A$ is not invertible, then the system has either:

  • No solutions
  • Infinite solutions parametrized by $t_0, t_1....$

By consequence since $det(A) = 0$ implies that $A$ is not a bijection. In terms of analysis, $A$ will be reduced to inconsistent if $b \notin \text{span}(A)$ where $\text{span}(A)$ is the linearly independent columns of $A$. I don't know of any other way to immediately tell for any arbitrary matrix.

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Since the matrix is not invertible, it has free variables. If you have $n-r=s$, where $n$ is the number of variables, and $r$ is the number of equations, $s$ is the amount of free variables.

If S $\neq $ 0 then the set has an infinite amount of solutions, and hence must be parameterized, is a theorem. It cannot be solved explicitly as you request. You would have to choose values for the free variables then a solution will follow.

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