solve this without using the determinant stuff, i didnt learn that yet

Certainly! Let's solve the problem using elementary row operations, which focus on simplifying the matrix to identify its rank as \( t \) varies.

### Given Matrix:
\[
A = \begin{pmatrix} 
1 & -2 & k \\ 
2 & 4t - 2 & 4t \\ 
k & 2 & 1 
\end{pmatrix}
\]

### Step 1: Perform Row Operations to Simplify the Matrix

We start by simplifying the matrix using row operations to make it easier to identify the rank.

#### 1. Eliminate the first entry in the second row.
Subtract 2 times the first row from the second row (R2):
\[
R2 = R2 - 2R1
\]
\[
\begin{pmatrix} 
1 & -2 & k \\ 
2 - 2 \times 1 & (4t - 2) - 2(-2) & 4t - 2k \\ 
k & 2 & 1 
\end{pmatrix}
=
\begin{pmatrix} 
1 & -2 & k \\ 
0 & 4t + 2 & 4t - 2k \\ 
k & 2 & 1 
\end{pmatrix}
\]

#### 2. Eliminate the first entry in the third row.
Subtract \( k \) times the first row from the third row (R3):
\[
R3 = R3 - kR1
\]
\[
\begin{pmatrix} 
1 & -2 & k \\ 
0 & 4t + 2 & 4t - 2k \\ 
k - k(1) & 2 + 2k & 1 - k^2 
\end{pmatrix}
=
\begin{pmatrix} 
1 & -2 & k \\ 
0 & 4t + 2 & 4t - 2k \\ 
0 & 2 + 2k & 1 - k^2 
\end{pmatrix}
\]

### Step 2: Analyze the Matrix to Determine Rank

At this stage, the matrix is in an easier form to analyze:

\[
\begin{pmatrix} 
1 & -2 & k \\ 
0 & 4t + 2 & 4t - 2k \\ 
0 & 2 + 2k & 1 - k^2 
\end{pmatrix}
\]

#### Case 1: \( t \) values where \( 4t + 2 \neq 0 \)

If \( 4t + 2 \neq 0 \), the second row has a leading non-zero entry, so the matrix can have full rank (rank = 3) if the third row also has a non-zero entry.

#### Case 2: \( t = -\frac{1}{2} \)

If \( t = -\frac{1}{2} \), then \( 4t + 2 = 0 \). The second row becomes:

\[
\begin{pmatrix} 0 & 0 & 4(-\frac{1}{2}) - 2k \end{pmatrix} = \begin{pmatrix} 0 & 0 & -2 - 2k \end{pmatrix}
\]

Now the matrix looks like:

\[
\begin{pmatrix} 
1 & -2 & k \\ 
0 & 0 & -2 - 2k \\ 
0 & 2 + 2k & 1 - k^2 
\end{pmatrix}
\]

The rank now depends on the third row:

- If \( k \) is such that both the second and third rows become zero, the rank of the matrix reduces further (rank = 1).
- If only one of these rows is non-zero, the rank is 2.

### Summary of Rank Conditions:

- **Rank 3:** This happens when \( t \neq -\frac{1}{2} \) and the third row is non-zero.
- **Rank 2:** This happens when \( t = -\frac{1}{2} \) and the third row is non-zero, or when the second row simplifies but the third row doesn't fully zero out.
- **Rank 1:** This occurs when both the second and third rows are fully zero, which requires specific conditions on \( k \) and \( t = -\frac{1}{2} \).

This approach avoids using determinants and uses row operations to understand the structure of the matrix as \( t \) varies.

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**Do you want further details on the specific values of \( k \) where the rank changes, or do you have other questions?**

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### Related Questions:
1. What happens to the rank if \( k = 0 \) and \( t = -\frac{1}{2} \)?
2. How do row operations generally help in identifying the rank of a matrix?
3. Why does the rank decrease when more rows become dependent?
4. What does it mean if the rank is 1 or 2 in practical terms?
5. How does the rank relate to the solutions of a system of equations represented by the matrix?

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**Tip:** Row operations are a powerful tool in linear algebra that can help simplify matrices and identify properties like rank without the need for determinants.

Learn how to determine the rank of a matrix using elementary row operations without using determinants. This problem involves analyzing a matrix with variable entries and understanding how the rank changes with different values of t and k. Suitable for college-level students studying linear algebra.

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