Question

The rank of the \( 4 \times 6 \) matrix \( \begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix} \) with entries in \( \mathbb{R} \), is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The Correct Option is D

To determine the rank of the given \(4 \times 6\) matrix:

1	1	0	0	0	0
1	0	1	0	0	0
0	1	0	1	0	1
0	0	1	0	1	1

1. Write down the matrix and apply row operations to obtain the row echelon form.
2. Convert the first two rows with leading coefficients '1' into pivot rows. They are already appropriate, with leading ones in different columns: row 1 and row 2 are pivot rows.
3. Considering row 3 (to include more pivot columns), perform the following operation to eliminate its leading zeros: \(R_3 \leftarrow R_3 - R_1\). After this operation, row 3 becomes:
   • \( (0, 0, 0, 1, 0, 1) \)
4. Perform a similar operation on row 4 to eliminate its leading ones: \(R_4 \leftarrow R_4 - R_2\). After this operation, row 4 remains unchanged as:
   • \( (0, 0, 0, 0, 1, 1) \)
5. Upon reaching row-echelon form, we identify pivot columns that contain the leading ones: columns 1, 2, 4, and 5.
6. Since these columns have leading ones, they contribute to the rank of the matrix.
7. The rank of a matrix is determined by the number of linearly independent rows (or pivot columns, equivalently), resulting in a rank of 4.

Conclusion: The rank of the matrix is 4