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$$3x+2y-z=4$$ $$4y-2z+6x=8$$ $$x-2y=5$$

I found that the first two equations were the same. Then, looking at the second and third equations, I thought the answer was No Solution, because I couldn't find a way to solve it, but instead the answer was Infinitely Many Solutions. Can someone explain this to me?

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  • $\begingroup$ "I couldn't find a way to solve it": there are zillion equations that have solutions though we don't know how to address them ! $\endgroup$ – Yves Daoust Sep 20 '16 at 12:49
  • $\begingroup$ Usually I can solve such problems though $\endgroup$ – thunderbolt Sep 21 '16 at 2:59
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As one of the equations is a duplicate, you can drop it and solve.

$$\begin{cases}3x+2y-z=4\\ x-2y=5\end{cases}$$

Let us move $z$ to the RHS and consider it as known to obtain a system of two equations in two unknowns,

$$\begin{cases}3x+2y=z+4\\ x-2y=5.\end{cases}$$

Then by elimination

$$\begin{cases}4x=z+9\\ 8y=z-11.\end{cases}$$

We can assign $z$ any value, yet get values of $x,y,z$ that satisfy all equations. For instance $(3,-1,3)$ or $(1,-2,-5)$. There are infinitely many others. In the solution space, $\mathbb R^3$, the solutions describe a straight line.

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This is the same as$$\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}4\\8\\5\end{bmatrix}$$

Since $\begin{bmatrix}3&2&-1\\6&4&-2\\4&-2&0\end{bmatrix}$ is non-invertable, it suggests 0 or infinite solutions.

Now we just have to find 1 solution to show it has infinite solutions.

Letting $y=0$, from $x-2y=5$, $x=5$, and then from $3x+2y-z=5$, $z=11$.

Generating solutions:

We can clearly see the first and second equations are identical, so we may forget about the second.

$$x=5+2y$$

Now plugging this into the first equation

$$z=3(5+2y)+2y-5=8y+10$$

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  • $\begingroup$ How do you know if a matrix in invertable? $\endgroup$ – thunderbolt Sep 15 '16 at 23:18
  • $\begingroup$ @thunderbolt It's determinant is 0, for this matrix it's easy to check, just 8 multiplications and 4 additions $\endgroup$ – Ariana Sep 15 '16 at 23:21
  • $\begingroup$ Ooh, ok, thanks. $\endgroup$ – thunderbolt Sep 15 '16 at 23:36
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The system depends on the arbitrary choice of one of the unknowns. Indeed, it is equivalent to $$\begin{cases}3x-2y-z=4\\x-2y=5\end{cases}\iff\begin{cases} x=5+2y\\z=-4+3x+2y=11+8y\end{cases}$$ So the set of solutions is $$\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix} 5\\0\\11\end{pmatrix}+y\begin{pmatrix} 2\\1\\8\end{pmatrix}.$$

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    $\begingroup$ Can you explain how you went from the second to the third step? Also, your notation is confusing to me; I only know how to write general solutions like $(2y+5,y,8y+11)$ $\endgroup$ – thunderbolt Sep 15 '16 at 23:16
  • $\begingroup$ It is just the matrix translation of your way to write your solution (your comment made me see I changed a sign in the equations. Problem is fixed) and correspond to the vector form of the parametric equation of a line: the origin on the line: $(5, 0,11)$ translated by a multiple of the direction vector: $(2,1,8)$. Is this clearer? $\endgroup$ – Bernard Sep 15 '16 at 23:24
  • $\begingroup$ I see it now. Thanks. $\endgroup$ – thunderbolt Sep 15 '16 at 23:35
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Let $z=t$ and removing the second equations, your problem reduced to

$$3x+2y=4+t$$ $$x-2y=5$$

Summing the two equations up, we have

$$4x=9+t$$

and we can conclude that $$x=\frac{9+t}{4}$$

Can you use $x=\frac{9+t}{4}$ and $x-2y=5$ to express $y$ in terms of $t$?

Whenever you fix a value of $t$, you obtain a different solutions of $(x,y,z)$, hence there are infinitely many solutions.

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  • $\begingroup$ Why do you use t? Why not keep it at z? $\endgroup$ – thunderbolt Sep 15 '16 at 23:16
  • $\begingroup$ That works too. $\endgroup$ – Siong Thye Goh Sep 16 '16 at 0:14

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