Question

If the ranks of the matrices \[ A= \begin{bmatrix} 1 & 0 & 1\\ 2 & 1 & 2\\ 1 & 0 & -1 \end{bmatrix} \] and \[ B= \begin{bmatrix} 1 & 2 & 3 & 4\\ 2 & 4 & 6 & -8 \end{bmatrix} \] are \(r_1\) and \(r_2\) respectively, then \(r_1-r_2=\)

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

For a square matrix, if the determinant is non-zero, then the rank is equal to the order of the matrix. Also, for a matrix with two rows, check whether the rows are scalar multiples to determine whether the rank is \(1\) or \(2\).

Answer 1

The Correct Option is B

Step 1: Set up the problem.
We need the ranks $ r_1 $ (rank of $ A $) and $ r_2 $ (rank of $ B $), then find $ r_1 - r_2 $. The rank of a matrix is the largest order of a non-zero minor.
Step 2: Compute the determinant of A.
\[ A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1 \end{bmatrix} \] Expanding along the first row (the middle entry 0 makes it easy): \[ |A| = 1 \cdot \begin{vmatrix} 1 & 2 \\ 0 & -1 \end{vmatrix} - 0 + 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} \] \[ = 1(1 \cdot (-1) - 2 \cdot 0) + 1(2 \cdot 0 - 1 \cdot 1) = (-1) + (-1) = -2 \]
Step 3: Determine rank of A.
Since $ |A| = -2 \neq 0 $, matrix $ A $ is invertible. A $ 3 \times 3 $ invertible matrix has rank 3. So $ r_1 = 3 $.
Step 4: Analyze matrix B.
$ B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & -8 \end{bmatrix} $. Compare rows: if $ R_2 $ were $ 2 R_1 $, we would get $ [2, 4, 6, 8] $. But the last entry of $ R_2 $ is $ -8 $, not $ 8 $. So $ R_2 \neq 2 R_1 $. The two rows are not proportional, meaning they are linearly independent.
Step 5: Determine rank of B.
Since the two rows of $ B $ are linearly independent, and $ B $ has only 2 rows, its rank is 2. So $ r_2 = 2 $.
Step 6: Compute the answer.
\[ r_1 - r_2 = 3 - 2 = 1 \] \[ \boxed{1} \]