Special Systems and Parameters

1. Homogeneous Systems & A Special Theorem

A linear system is considered homogeneous if all the constant terms (the numbers on the right side of the equals sign) are zeros.

Homogeneous systems will never have "no solution" because you can always just plug in $0$ for all variables (e.g., $x = 0, y = 0$).

Important Theorem

A homogeneous system with more unknowns than the number of equations always has infinitely many solutions.

Why? If you have 4 variables (unknowns) but only 3 equations, you won't have enough "information" to lock down a single point. You will inevitably have to use a parameter (like $t$), which creates an infinite number of solutions.

2. Advanced Gauss-Jordan: Free Variables and Parameters

Sometimes, when reducing a matrix, a row will completely zero out on the left side, meaning you don't have a "leading 1" for every variable. When this happens, you have infinitely many solutions and must use parameters.

Complex Example (5 variables)

Imagine you use Gauss-Jordan to reduce a 3-equation, 5-variable system to this RREF:

$$\left[\begin{array}{ccccc|c} 1 & -1 & 0 & 2 & -1 & 2 \\ 0 & 1 & -2 & 2 & -1 & 7 \\ 0 & 0 & 1 & -\frac{2}{3} & \frac{4}{3} & -\frac{8}{3} \end{array}\right]$$

Notice that variables $x_1, x_2,$ and $x_3$ have leading 1s, but $x_4$ and $x_5$ do not. The variables without leading 1s become our parameters. Let's assign them:

• Let $x_4 = s$
• Let $x_5 = t$

Now, write out the equations from the matrix and substitute $s$ and $t$:

This methodical back-substitution is how you solve massive systems with multiple free variables!

3. Analyzing Systems with Unknown Constants ($k$)

A classic exam problem asks you to determine what values of a constant $k$ will result in no solutions, one solution, or infinite solutions.

The Strategy

Reduce the matrix until you get an expression with $k$ on the bottom row. Let's look at an advanced example from your notes.

After several row operations, your augmented matrix reduces to:

$$\left[\begin{array}{ccc|c} 1 & 1 & 3 & 2 \\ 0 & k+2 & -1 & -1 \\ 0 & 0 & k^2-21 & k-11 \end{array}\right]$$

Let's analyze the bottom row, which translates to the equation: $(k^2 - 21)z = (k - 11)$.

Finding "No Solution"

We look for a contradiction where $0z = \text{non-zero}$.

Case 1: If $k^2 - 21 = 0$, then $k = \sqrt{21}$ or $k = -\sqrt{21}$.

If we plug in $\sqrt{21}$, the equation becomes $0z = \sqrt{21} - 11$ (which is a non-zero number). This is a contradiction, so there are no solutions at $k = \pm\sqrt{21}$.

Case 2 (Hidden trap!): What if $k = -2$?

Look at the second row: if $k = -2$, the second row becomes $0y - z = -1$ (so $z = 1$). But the third row becomes $(4 - 21)z = -2 - 11$, which simplifies to $-17z = -13$ (so $z = \frac{13}{17}$). Since $z$ cannot be both $1$ and $\frac{13}{17}$ at the same time, this is a contradiction.

Answer for No Solution: $k \in \{-2, \sqrt{21}, -\sqrt{21}\}$

Finding "Exactly One Solution"

This happens when we avoid all the contradictions above. As long as $k$ doesn't cause a division by zero or a logical contradiction, we will find a single exact point.

Answer for One Solution: $k \in \mathbb{R} \setminus \{-2, \sqrt{21}, -\sqrt{21}\}$