Solving three equations

Question

I was just wondering how would one solve three equations which are in a form like these. Tried substitution, elimination and everything but couldn't find out the solution.

The equations are : $-2A+C=3$, $3A+B-2C=1$, $-A+C-B=-2$

Edit- Whenever I try to solve two equations, I arrive at an already provided third equation. Does that mean that these equations are true for more than one values of $A$, $B$ and $C$(they are arbitrary)?

Answer 1

Substitution and elimination will work as normal but you need to work with all three in each step.

Using Substitution

Rearrange first: $C=3+2A$

Sub into the other two: $3A+B-2(3+2A)=1$ and $-A+3+2A-B=-2$

Simplifies to: $-A+B=7$ and $A-B=-5$

Then rearrange the first of these: $B=7+A$

Sub int the other: $A-(7+A)=-5$

As $-7\neq-5$ then there is no solution.

Reason for no solution

As you know from solving two equations you get no solution if the two lines are parallel. Each of your equations represents a plane in 3D space. You have the special situation where the three planes do not intersect at a single point.

Update:

Just read your edit. If you arrived at the third equation using the first two then you made a mistake in your arithmetic. If I change the question slightly we can get that situation.

Modified equations: $−2A+C=3$, $3A+B−2C=\color{red}{-1}$, $−A+C−B=−2$

Using Substitution

Rearrange first: $C=3+2A$

Sub into the other two: $3A+B-2(3+2A)=-1$ and $-A+3+2A-B=-2$

Simplifies to: $-A+B=5$ and $A-B=-5$

Then rearrange the first of these: $B=5+A$

Sub int the other: $A-(5+A)=-5$

As $5=-5$ then there is an infinite number of solutions.

We could let $A=t, t\in\mathbb{R}$ and hence $B=5+t$ and hence $C=3+2t$

Graphically the three planes would be like this. Different values of $t$ represents different points long the line of intersection.

Answer 2

Working with the augmented matrix: \begin{align*} \left[\begin{array}{rrr|r} -2&0&1&3\\3&1&-2&1\\-1&-1&1&-2 \end{array}\right]&\rightsquigarrow \left[\begin{array}{rrr|r} 1&1&-1&2\\-2&0&1&3\\3&1&-2&1 \end{array}\right]\rightsquigarrow \left[\begin{array}{rrr|r} 1&1&-1&2\\0&2&-1&7\\0&-2&1&-5 \end{array}\right]\\&\rightsquigarrow \left[\begin{array}{rrr|r} 1&1&-1&2\\0&2&-1&7\\0&0&0&2 \end{array}\right] \end{align*} The rank of the augmented matrix is greater than the rank of the left-hand matrix, hence the system has no solution.