How can you identify how many solutions systems have in $\mathbb{R}^3$?

Question

How to determine how many solutions does each of the following systems have in $\mathbb{R}^3$? (Infinitely many solutions/Unique solutions/No Solutions)

a) $\begin{cases}f + g + h = 13\\ f – h = −2\end{cases}$

b) $\begin{cases}3x + 4y – z = 8\\ 5x – 2y + z = 4\\ 2x – 2y + z = 1\end{cases}$

c) $\begin{cases}S – T – W = 8\\ 5S + 2T + 4W = 0\\ –3S + 3T + 3W =20\end{cases}$

Answer 1

For (c), adding three times the first equation to the third equation yields $0=44$, which is clearly impossible, so there are no solutions.

For (b), subtracting the third equation from the second yields $3x=3$ or $x=1$; thus $3+4y-z=8$ and $5-2y+z=4$; adding those yields $8+2y=12$ or $y=2$ and thus $z=3$, so $(x,y,z)=(1,2,3)$ is the unique solution.

For (a), adding and subtracting the equations we get $2f+g=11$ and $h=f+2$; $f$ could be anything, and then $g$ and $h$ are determined, so there are infinitely many solutions.

Answer 2

You can use the determinant to evaluate whether the system has unique solutions.

• $ det A \neq 0 \iff \text{unique solution}$
• $ det A = 0 \implies \text{infinite number of solutions or none}$

The simplest way to determine between infinite or none is to solve and see if the system is inconsistent (results in contradictory constraints).

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