Question: Determine which values of \(k\), if any, will give
(a) a unique solution, (b) no solution, (c) infinitely many solutions for:
$$
\begin{aligned}
x &+ 2y &- z &= 1\\
2x &+ 3y &+ z &= 1\\
-4x &- 5y &+ (k^2-9)z &= k+1
\end{aligned}
$$
Step 1: Write the augmented matrix and row-reduce (REF)
Start with the augmented matrix \([A\mid b]\):
$$
\left[\begin{array}{ccc|c}
1 & 2 & -1 & 1\\
2 & 3 & 1 & 1\\
-4 & -5 & k^2-9 & k+1
\end{array}\right]
$$
After row reduction to REF, we reach:
$$
\left[\begin{array}{ccc|c}
1 & 2 & -1 & 1\\
0 & 1 & -3 & 1\\
0 & 0 & k^2 – 4 & k+2
\end{array}\right]
$$
The last row represents:
$$
(k^2-4)z = k+2
$$
Step 2: Use the last row to decide the solution type
(a) Unique solution
A unique solution happens when the last pivot entry is nonzero:
$$
k^2-4 \neq 0 \;\Rightarrow\; k \neq 2,\,-2
$$
✅ Unique solution: \(k \neq 2,\,-2\)
(b) No solution
No solution happens when we get a contradiction \(0=\text{nonzero}\).
This occurs when:
$$
k^2-4 = 0 \quad \text{and} \quad k+2 \neq 0
$$
If \(k=2\), the last row becomes:
$$
0\cdot z = 4 \;\Rightarrow\; 0=4
$$
❌ No solution: \(k=2\)
(c) Infinitely many solutions
Infinitely many solutions happen when the last row becomes \(0=0\).
If \(k=-2\), the last row becomes:
$$
0\cdot z = 0
$$
So one variable is free ⇒ infinitely many solutions.
♾️ Infinitely many solutions: \(k=-2\)
Final Answer
- Unique solution: \(k\neq 2,\,-2\)
- No solution: \(k=2\)
- Infinitely many solutions: \(k=-2\)
Quick checklist (10 seconds)
After you reach REF, look only at the last row:
$$
[0\ \ 0\ \ k^2-4 \mid k+2]
$$
- If \(k^2-4\neq 0\) ⇒ pivot in \(z\) ⇒ unique solution
- If \(k^2-4=0\) and \(k+2\neq 0\) ⇒ \(0=\text{nonzero}\) ⇒ no solution
- If \(k^2-4=0\) and \(k+2=0\) ⇒ \(0=0\) ⇒ infinitely many solutions
Practice (based on this example)
1) If \(k=3\), what type of solution do we get?
2) If \(k=2\), what type of solution do we get?
3) If \(k=-2\), what type of solution do we get?
Answers
1) \(k=3\): unique solution
2) \(k=2\): no solution
3) \(k=-2\): infinitely many solutions
Extra Practice (2 variables: \(x, y\))
1) Classify the solution type
Determine for what values of \(k\) the system has a unique solution, no solution, or infinitely many solutions:
$$
\begin{aligned}
x + 2y &= 3\\
2x + 4y &= k
\end{aligned}
$$
Answers
- If \(k=6\) ⇒ infinitely many solutions
- If \(k\neq 6\) ⇒ no solution
- Unique solution: none
2) When is there a unique solution?
Determine for what values of \(k\) the system has a unique solution:
$$
\begin{aligned}
kx + y &= 1\\
x + y &= 2
\end{aligned}
$$
Answers
- If \(k\neq 1\) ⇒ unique solution
- If \(k=1\) ⇒ no solution
- Infinitely many solutions: none