Linear System of Equations

Question

I've been given: $$ \begin{array}{rcrcrcr} x & + &2y &+& z &=& 3\\ x & +& y &+& 2z &=& 2\\ & & y&-& z &=& 1 \end{array} $$ Let $S$ be the set of all triples that are solutions to the system above and $$T = {(x,y,z)}={(1-3t, 1+t, t)} $$ I need to show that $S=T$.

So, I tried solving for the system of equations first which got me nowhere because solving the first two equations gives us the third equation. So that was a bust.

Also, I'm not sure what to do with T?

Answer 1

The form of the triples in $T$ suggests setting $z=t$.

Then the first two equations become $x + 2y = 3 - t$ and $x + y = 2 - 2t$.

Solving, we get $x = 1 - 3t$ and $y = 1 + t$, as expected.

Thus, any solution to the system is in $T$, and $S \subseteq T$.

It's easy to verify any triple in $T$ also satisfies the system, and is thus in $S$, so $T \subseteq S$.

Therefore, $S = T$.

Answer 2

The system does not have a unique solution precisely because these equations are not independent. In other words, writing it as $$\underbrace{\begin{bmatrix}1 & 2 & 1 \\1 & 2 & 2 \\ 0 &1 & -1 \end{bmatrix}}_{=A}\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} 3 \\ 2 \\ 1\end{bmatrix},$$the matrix $A$ does not have full rank. The solution of that system is the kernel of $A$, shifted by a particular solution of that system (for example, $(1,1,0)$). You'll have a degree of freedom.

Without appealing to matrices, you can check that every element of the form $(1-3t,1+t,t)$ is a solution of that system (plug it!), which says $T \subseteq S$. On the other hand, you can solve for $x$ and $y$ in terms of $z$, and call $z = t$. You'll get precisely an element of the form $(1-3t,1+t,t)$, which tell us that $S \subseteq T$.

Answer 3

• First of all, do you know how to solve a linear equation by Gaussian elimination?

• To show $T=S$, you could show $T\subset S$ and $S\subset T$. One of the direction is trivial, can you see why?

• Note that $$ T = \{(1-3t, 1+t, t) \, | \, t \in \mathbb{R}\}=\{u+tv\mid t\in\mathbb{R}\} $$ where $u=(1,1,0)$ and $v=(-3,1,1)$. Can you see why you should be expecting that the solution space to $Ax=0$ where $A$ is the matrix of your linear equation should be of dimension $1?$

• Do you know how to systematically solve $Ax=0$?