Question

The rank of the matrix \[ \begin{bmatrix} -4 & -1 & 1 & 4 \\ -3 & 0 & 2 & 3 \\ -2 & -1 & 0 & -4 \end{bmatrix} \] is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For a $3\times4$ matrix, if any $3\times3$ minor is non-zero, the rank is immediately 3.

Answer 1

The Correct Option is D

Concept:
The rank of a matrix is the maximum number of linearly independent rows or columns. To determine the rank, we transform the matrix into row-echelon form using elementary row operations. 

Step 1: Write the matrix.
\[ A= \begin{bmatrix} -4 & -1 & 1 & 4 \\ -3 & 0 & 2 & 3 \\ -2 & -1 & 0 & -4 \end{bmatrix}. \] 

Step 2: Apply row operations.
Interchanging rows and simplifying, \[ R_2 \to 4R_2 - 3R_1 \] gives \[ \begin{bmatrix} -4 & -1 & 1 & 4 \\ 0 & 3 & 5 & 0 \\ -2 & -1 & 0 & -4 \end{bmatrix}. \] Now \[ R_3 \to 2R_3 - R_1 \] gives \[ \begin{bmatrix} -4 & -1 & 1 & 4 \\ 0 & 3 & 5 & 0 \\ 0 & -1 & -1 & -12 \end{bmatrix}. \] 

Step 3: Check non-zero rows.
The determinant of the leading $3\times3$ minor is \[ \begin{vmatrix} -4 & -1 & 1 \\ 0 & 3 & 5 \\ 0 & -1 & -1 \end{vmatrix} = -4(-3 + 5) = -8 \neq 0. \] Thus a non-zero $3\times3$ minor exists. Therefore, \[ \operatorname{Rank}(A) = 3. \] Hence, \[ \boxed{3} \]