1. ## Matrix equation proof

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

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

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

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

4. ## Here´s what I got so far

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

means that $\displaystyle X_3$ represents the multitude of solutions.Right?
And if i rewrite the original equation to $\displaystyle A\left(sX_1+tX_2\right)=B$ and prove that it´s true for any $\displaystyle (s,t\in R)$ then im done. Right?

I tried this:
$\displaystyle A\left(sX_1+tX_2\right)=B$

$\displaystyle AsX_1+AtX_2=B$

$\displaystyle sAX_1+tAX_2=B$

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

And since $\displaystyle AX_1=AX_2$

I get

$\displaystyle sAX_1+tAX_1=B$

$\displaystyle (s+t)AX_1=B$

I can rewrite the equation $\displaystyle sAX_1+tAX_2=B$ to $\displaystyle sB+tB=(s+t)B=B$

because $\displaystyle AX_1=AX_2=B$

The final equation looks like this

$\displaystyle (s+t)AX_1=(s+t)B$ which is true for any $\displaystyle (s,t\in R)$ since $\displaystyle (s+t)$ 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 $\displaystyle s+t$ because $\displaystyle X_1$ is a solution.
You can try to find $\displaystyle s$ and $\displaystyle t$ such as $\displaystyle sX_1+tX_2$ is a solution given that $\displaystyle X_1$ and $\displaystyle X_2$ 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 $\displaystyle X_1$ and $\displaystyle X_2$ are solutions of $\displaystyle AX=B$ then what can you say about $\displaystyle t X_1+\left(1-t\right)X_2$ for $\displaystyle t\in \mathbb R$?

8. ## Still confused

I should probably give up since I still do not understand. I´m assuming that $\displaystyle X_1$ and $\displaystyle X_2$ 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

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

..nothing really!

9. You can multiplate it by $\displaystyle A$.