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Consider the matrix $A$, to be equal to: \begin{bmatrix}1&2&1\\-1&4&3\\2&-2&a\end{bmatrix}

Then we can rewrite this as: \begin{bmatrix}1&2&1\\0&6&4\\2&-2&a\end{bmatrix} \begin{bmatrix}1&2&1\\0&1&2/3\\2&-2&a\end{bmatrix} \begin{bmatrix}1&2&1\\0&1&2/3\\0&-6&a-2\end{bmatrix} \begin{bmatrix}1&2&1\\0&1&2/3\\0&0&a+2\end{bmatrix}

Now consider the system $Ax=0$. If $\mathbf{a = -2}$, then $x_3$ is a free variable, because a+2 turns to zero. The solution becomes: $$x_1 = (1/3)x_3$$ $$x_2 = -(2/3)x_3$$ $$x_3 = free$$ If $\mathbf{a}$ does not equal 2 , then the system seems to have trivial solutions every time, because you divide $a+2$ by itself, which becomes $1$, regardless of the value of $a$. The solutions become: $$x_1 = 0$$ $$x_2 = 0$$ $$x_3 = 0$$

But then how can you determine when the system $Ax=0$ is inconsistent or has infinitely many solutions? What mistake have I made? Could someone provide me your own method if my method seems to be wrong?

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2 Answers 2

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We are talking of the homogeneous system of equations, right?

The solution for $a=-2$ is correct. As you have said, for $a\neq -2$, the only solution is $x_1=x_2=x_3=0$.

The dimension of the solution space is thus $0$ if $a=-2$ and $1$ otherwise.

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    $\begingroup$ OP is talking about infinitely many solutions, not infinitely many independent solutions $\endgroup$ Commented Mar 24, 2019 at 11:59
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    $\begingroup$ @Stallmp There is always at least one solution for homogeneous systems of equations (the trivial solution $0$); there are infinitely many solutions for $a=-2$. $\endgroup$ Commented Mar 24, 2019 at 12:02
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    $\begingroup$ @Stallmp Yes, there is always a solution. $\endgroup$ Commented Mar 24, 2019 at 12:04
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    $\begingroup$ @Stallmp Your calculations are correct. $\endgroup$ Commented Mar 24, 2019 at 12:06
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    $\begingroup$ @Stallmp: When $a=-2$, as you wrote, you have as solutions $S = \{ (v/3, -2v/3, v): v \in \mathbb{R} \}$. $\endgroup$ Commented Mar 24, 2019 at 12:07
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$Ax=0$ has a unique solution if $\det(A)\neq 0$ and it has infintely many solutions if $\det(A)= 0$ (any multiple of a solution is also a solution). In this case $\det(A)=0$ iff $a =-2$.

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  • $\begingroup$ Alright I see, thank you. Would this imply that inconsistency is not possible with systems Ax = 0 ? $\endgroup$ Commented Mar 24, 2019 at 12:07
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    $\begingroup$ Yes, the zero vector is always a solution, so there is no question of inconsistency. $\endgroup$ Commented Mar 24, 2019 at 12:09
  • $\begingroup$ Alright thanks a lot! $\endgroup$ Commented Mar 24, 2019 at 12:09

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