find the value of k so that the rank of the following matrix is 3 [[6,3,5,9],[5,3,2,6],[3,1,2,6],[2,1,1,k]]

We are tasked with finding the value of \( k \) such that the rank of the matrix is 3. The matrix is:

\[
A = \begin{bmatrix}
6 & 3 & 5 & 9 \\
5 & 3 & 2 & 6 \\
3 & 1 & 2 & 6 \\
2 & 1 & 1 & k
\end{bmatrix}
\]

### Step-by-step process:

1. **Construct an augmented matrix** and perform row operations to put it into echelon form.
2. The rank of the matrix is the number of non-zero rows after row reduction.
3. Set \( k \) such that the rank is 3, meaning the determinant of any 4x4 submatrix should be 0 (since the full matrix would be rank-deficient).

I'll compute the determinant of the 4x4 matrix and set it equal to zero to solve for \( k \).The determinant of the given matrix is \( 24 - 8k \). To ensure the rank of the matrix is 3, we set the determinant to 0:

\[
24 - 8k = 0
\]

Solving for \( k \):

\[
k = 3
\]

Thus, the value of \( k \) that makes the rank of the matrix equal to 3 is \( \boxed{3} \).

Would you like further details on the row-reduction process or any other clarification?

Here are some related questions:
1. What does matrix rank represent in linear algebra?
2. How is the determinant used to assess matrix rank?
3. How does row reduction work to simplify matrices?
4. What happens if the matrix's rank is less than 3?
5. Can matrix rank be affected by slight changes in matrix elements?

**Tip:** The rank of a matrix can be used to determine the number of linearly independent rows or columns it contains.

Discover how to find the value of k such that the rank of a 4x4 matrix is 3. This step-by-step solution involves key linear algebra concepts like determinants, row reduction, and the Rank-Nullity Theorem. Suitable for undergraduate students studying linear algebra.

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