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$$\left(\begin{array}{cccc} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0 \end{array}\;\middle\vert\;\begin{array}{c}2\\3\\4\\0\end{array}\right)$$

This is a $4$ by $4$ matrix $A$ and the solutions in $\mathbb{R}^4$ $(2, 3, 4, 0)$.

I think the lowest row makes $x_4$ free variable so this linear system has infinite solutions. Correct?


$$\left(\begin{array}{ccc} 1&0&0\\0&1&0\\0&0&1\\0&0&0 \end{array}\;\middle\vert\;\begin{array}{c}2\\3\\4\\0\end{array}\right)$$

This is a $4$ by $3$ matrix so four equations and three variables. This time this linear system has one unique solution right? No free variable? Please let me know if I am on the right track.

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  • $\begingroup$ Yes, you are. You can also easily write all solutions of the first system: $x_4$ can be anything. $\endgroup$ – egreg Apr 23 '13 at 22:58
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$(1)$ Yes, you are correct. $x_4$ is a "free variable", and there are indeed an infinite number of solutions to the given system of equations - not just the particular vector you've chosen by letting $x_4 = 0$. That is one solution, but there is a family of solutions, which is infinite in number. That's not to say every vector is a solution: the solutions simply depend on $x_4$, and there are infinitely many choices for $x_4 \in \mathbb R$.

You can represent these of solution-vectors (all solutions where $x, y, z$ are determined, and vary only by the choice of $x_4 \in \mathbb R)$, by using a parameter, $\color{blue}{\bf x_4 = t}$ where $t\in \mathbb R$, as follows:

$$\begin{pmatrix} 2 \\ 3 \\ 4 \\ \color{blue}{\bf t} \end{pmatrix}\,\in\, \mathbb R^4, \quad t\in \mathbb R$$

$(2)$ Your conclusions about the solution to your second problem is correct: you have a unique solution given by:

$$\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} \in \mathbb R^3$$

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for example if you solve system

$x_1+x_2=1$

$x_1-2x_2=0$

your answer in $R^2$is only $(\frac{2}{3},\frac{1}{3})$

but in $R^{3}$

$(\frac{2}{3},\frac{1}{3},x_3)$ $x_3 \in R$ is solution you also : your first system solved in $R^4$ and second solved in $R^3$

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