Question

The rank of matrix \( \begin{bmatrix} 1 & 2 & 2 & 3 \\ 3 & 4 & 2 & 5 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{bmatrix} \) is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

To determine the rank of a matrix, row reduce it to row echelon form and count the number of non-zero rows.

Answer 1

The Correct Option is B

To determine the rank of a matrix, we can use row reduction to bring the matrix into row echelon form (REF). The rank is the number of non-zero rows in the REF. Let's apply row reduction to the matrix: \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 3 & 4 & 2 & 5 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{bmatrix} \] Step 1: Subtract 3 times the first row from the second row: \[ R_2 = R_2 - 3R_1 \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 5 & 6 & 2 & 7 \\ 7 & 8 & 2 & 9 \end{bmatrix} \] Step 2: Subtract 5 times the first row from the third row: \[ R_3 = R_3 - 5R_1 \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 7 & 8 & 2 & 9 \end{bmatrix} \] Step 3: Subtract 7 times the first row from the fourth row: \[ R_4 = R_4 - 7R_1 \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & -2 & -4 & -4 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{bmatrix} \] Step 4: Divide the second row by -2: \[ R_2 = \frac{R_2}{-2} \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & -4 & -8 & -8 \\ 0 & -6 & -12 & -12 \end{bmatrix} \] Step 5: Add 4 times the second row to the third row: \[ R_3 = R_3 + 4R_2 \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & -6 & -12 & -12 \end{bmatrix} \] Step 6: Add 6 times the second row to the fourth row: \[ R_4 = R_4 + 6R_2 \] \[ A = \begin{bmatrix} 1 & 2 & 2 & 3 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \] Now, we see that there are two non-zero rows, so the rank of the matrix is 2. Therefore, the correct answer is option (B). \\ Final Answer: (B) 2