1. ## Matrix equation proof

If a matrix equation $AX=B$ has two solutions then why does it have infinitely many solutions?

Clue: if $X_1$ and $X_2$ are solutions then look into $X_3=sX_1+tX_2 (s,t\in R)$

Now, I really don´t know where to start on this one. I would appreciate some help just to get started.

2. Take s and t such that $s+t=1$. Let $X_1, X_2$ be such that $AX_1 = AX_2 = B$. Now, what can you say about $X_3 = s \cdot X_1 + t \cdot X_2$ ?

3. I guess the unknown is $X$.
If $X_1$ and $X_2$ are solutions, we will have $A\left(sX_1+tX_2\right) = sB+tS =\left(s+t\right)B$ and if $s+t=1$ then $sX_1+tX_2$ will be a solution, that can be different from $X_1$ and $X_2$ for some $s$ and $t$ well chosen.

4. ## Here´s what I got so far

This line $

X_3=sX_1+tX_2 (s,t\in R)
$

means that $X_3$ represents the multitude of solutions.Right?
And if i rewrite the original equation to $

A\left(sX_1+tX_2\right)=B
$
and prove that it´s true for any $(s,t\in R)$ then im done. Right?

I tried this:
$

A\left(sX_1+tX_2\right)=B
$

$AsX_1+AtX_2=B$

$sAX_1+tAX_2=B$

Now I can do that right? Because $s$ and $t$ are scalars.

And since $AX_1=AX_2$

I get

$sAX_1+tAX_1=B$

$(s+t)AX_1=B$

I can rewrite the equation $

sAX_1+tAX_2=B
$
to $

sB+tB=(s+t)B=B
$

because $

AX_1=AX_2=B
$

The final equation looks like this

$

(s+t)AX_1=(s+t)B
$
which is true for any $(s,t\in R)$ since $(s+t)$ is a scalar and can be divided from the equation.

Honestly, if this is correct I´m still unsure so any enlightment is still appreciated.

5. You could get this equation more directly by multiplying by $s+t$ because $X_1$ is a solution.
You can try to find $s$ and $t$ such as $sX_1+tX_2$ is a solution given that $X_1$ and $X_2$ are solutions.

6. I´m not quite sure if I understand. I see that i could cut to the chase a lot faster but I guess it´s the "proof" part that I tried to elaborate from the clue-part. Can I consider your first reply an answer to the problem?

I´m having a hard time grasping the s and t part.

And is there something wrong about my attempt to an answer besides it being unnecessarily long?

7. If $X_1$ and $X_2$ are solutions of $AX=B$ then what can you say about $t X_1+\left(1-t\right)X_2$ for $t\in \mathbb R$?

8. ## Still confused

I should probably give up since I still do not understand. I´m assuming that $X_1$ and $X_2$ are two different matrixes that satisfy the equation. What´s the purpose of adding them together? Let alone the purpose of adding the for some well chosen values of s and t?

So what can I say about

$

t X_1+\left(1-t\right)X_2
$

..nothing really!

9. You can multiplate it by $A$.