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I learnt that for the equation: $Ax=b$

There is one solution if $A$ is invertible.

But if $A$ is singular there are infinity solutions or no solutions at all.

If $A$ is singular is it possible to determine whether there are no solutions or infinity solutions?

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4 Answers 4

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Yes, the general theory says that the solutions exist iff

$$ rk(A)=rk(\bar{A}), $$

where $\bar{A}$ is the matrix with the additional column $b$ appended to it (it is called the augmented matrix).

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  • $\begingroup$ Why the downvote? Isn't this correct? $\endgroup$
    – Vossler
    Mar 31, 2016 at 14:46
  • $\begingroup$ To me the Question is how-to-tell, not yes or no. I'm not the down voter however. $\endgroup$
    – hardmath
    Mar 31, 2016 at 14:51
  • $\begingroup$ I have no idea - looks correct to me! $\endgroup$
    – almagest
    Mar 31, 2016 at 14:51
  • $\begingroup$ @hardmath I thought this answers the how-to-tell question: compare the ranks. Whatever, though. $\endgroup$
    – Vossler
    Mar 31, 2016 at 14:54
  • $\begingroup$ Sure, that is true. I'd say a few words about row reduction but that's only one perspective. $\endgroup$
    – hardmath
    Mar 31, 2016 at 14:55
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Yes, to examine the size of the solution set of a system of linear equations, we look at the rank of the coefficient matrix compared with the rank of the augmented matrix.

Let $A$ be the coefficient matrix of a system of linear equations in $n$ unknowns and $\mathbf{b}$ be the vector of constants. If $\text{rank}(A|\mathbf{b}) = \text{rank}(A) = n$ then we have a single unique solution. If you put a square matrix into reduced row-echelon form and it has rank $n$ then the reduced row-echelon form of that matrix is $\mathbb{I}_n$ the identity matrix, which is equivalent to having a unique solution to the system.

If $\text{rank}(A|\mathbf{b}) = \text{rank}(A) < n$ then there are infinitely many solutions to the system. In this case, we see that the row-echelon form of the matrix has a row of zeroes at the bottom and this means that at least one of the variables is a $\textit{free variable}$.

If $\text{rank}(A|\mathbf{b}) > \text{rank}(A)$ then there are no solutions. This is easy to see, since there will be a row of zeroes in the row echelon form of $A$ but a row of the form $[0\ 0 \ 0 \cdot\cdot\cdot\ c]$ in $(A|\mathbf{b})$ with $c \neq 0$, which corresponds to a linear equation of the form

$$0a_1 + 0a_2 + 0a_3 + \dots + 0a_n = c$$

which, of course, is nonsense.

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  • $\begingroup$ Not to discredit your answer in any way, but do you have any idea why was my answer downvoted? Isn't it the same thing? $\endgroup$
    – Vossler
    Mar 31, 2016 at 14:51
  • $\begingroup$ I agree with the comments below your post. Perhaps the lack of detail was the reason for the downvote. I am unsure why people are downvoting it! $\endgroup$ Mar 31, 2016 at 14:52
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A simple and fast answer is that if $b$ is in the space spanned by the column of $A$, there will be infinite solutions. Otherwise, there is no solution.

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Your question's title isn't logical, but see Proposition 3.38 in Kuldeep Singh's Linear Algebra: Step by Step (2013) on p. 265. Typo is "Exercises 3.6" that ought "Exercise 3.6.14" on p 268.

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