A=[1,2,0,a;b,0,1,c;0,d,1,0]. Find the rank of A based on the values a, b, c and d.

To find the rank of the matrix \( A \) based on the values of \( a, b, c, \) and \( d \), we need to analyze how these variables affect the linear independence of the rows (or columns) of \( A \). The matrix \( A \) is given by:

\[
A = \begin{bmatrix}
1 & 2 & 0 & a \\
b & 0 & 1 & c \\
0 & d & 1 & 0
\end{bmatrix}
\]

### Step-by-Step Determination of Rank

1. **Initial Matrix Analysis:**
   The rank of a matrix is the maximum number of linearly independent rows (or columns) in the matrix. To determine this, we'll perform Gaussian elimination to reduce the matrix to row echelon form (REF).

2. **Row Operations:**
   Let's start performing row operations to simplify the matrix:
   
   - The first row is already fine with a leading 1.
   - For the second row, subtract \(b\) times the first row from the second row to eliminate \(b\) in the first column:
   
   \[
   R_2 = R_2 - bR_1 \implies \begin{bmatrix}
   1 & 2 & 0 & a \\
   0 & -2b & 1 & c - ab \\
   0 & d & 1 & 0
   \end{bmatrix}
   \]

   - Next, for the third row, we don't need to change anything just yet since it already has a zero in the first column.
   
3. **Further Simplification:**
   Now, let’s focus on eliminating elements below the pivot in the second column. We add \(\frac{d}{-2b}\) times the second row to the third row:
   
   \[
   R_3 = R_3 + \frac{d}{-2b}R_2 \implies \begin{bmatrix}
   1 & 2 & 0 & a \\
   0 & -2b & 1 & c - ab \\
   0 & 0 & 1 + \frac{d}{2b} & \frac{d(c - ab)}{-2b}
   \end{bmatrix}
   \]

4. **Determining the Rank:**
   - **Rank 3:** If all rows are linearly independent, the rank is 3. This happens if:
     \[
     -2b \neq 0, \quad 1 + \frac{d}{2b} \neq 0.
     \]
     The first condition gives \( b \neq 0 \). The second condition simplifies to \( 1 + \frac{d}{2b} \neq 0 \), or equivalently, \( d \neq -2b \).

   - **Rank 2:** If the third row becomes a linear combination of the first two, or if the second row is zeroed out, the rank is 2. This happens if:
     \[
     b = 0 \quad \text{or} \quad d = -2b.
     \]

     Specifically:
     - If \( b = 0 \), the second row becomes \([0, 0, 1, c]\), and depending on \( c \), the third row could provide an additional pivot.
     - If \( d = -2b \) but \( b \neq 0 \), then the condition \( 1 + \frac{d}{2b} = 0 \) causes the third row to be zeroed out.

   - **Rank 1:** For rank 1, both the second and third rows would need to be combinations of the first or zero rows, which only happens if \( b = 0 \), \( c = 0 \), and \( d = 0 \).

### Conclusion

- **Rank 3:** If \( b \neq 0 \) and \( d \neq -2b \).
- **Rank 2:** If \( b = 0 \) (regardless of \( d \)) or \( d = -2b \) (with \( b \neq 0 \)).
- **Rank 1:** If \( b = 0 \), \( c = 0 \), and \( d = 0 \).

Would you like more details on any of these steps or have any questions?

### Related Questions:

1. What would be the rank of matrix \( A \) if \( a = 0, b = 1, c = 1, \) and \( d = -2 \)?
2. How does the determinant of matrix \( A \) change with different values of \( a, b, c, \) and \( d \)?
3. How would we perform row operations if \( a, b, c, \) and \( d \) were given specific numeric values?
4. How does changing the order of rows affect the row echelon form of \( A \)?
5. Can the matrix \( A \) be invertible, and under what conditions?

### Tip:
Always start with row operations that create zeros in the lower triangular part to simplify finding the rank.

Learn how to determine the rank of matrix A based on the values of variables a, b, c, and d. Explore the step-by-step process using Gaussian elimination to simplify the matrix and identify linear independence. Suitable for advanced mathematics students and professionals.

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