Question

The rank of the matrix \[ A= \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix} \] is

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

Hint:

For a \(3\times 3\) matrix, if the determinant is zero, the rank is less than \(3\). Then check whether any \(2\times 2\) minor is non-zero. If yes, the rank is \(2\).

Answer 1

The Correct Option is C

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

Step 2: Check whether the determinant is zero.
If the determinant of a \(3\times 3\) matrix is non-zero, then its rank is \(3\). If the determinant is zero, then its rank is less than \(3\).
Now, \[ |A|= \begin{vmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{vmatrix} \] Expanding along the first row, \[ |A| = 2\begin{vmatrix} 0 & 1 \\ 1 & 4 \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 4 & 4 \end{vmatrix} + 2\begin{vmatrix} 1 & 0 \\ 4 & 1 \end{vmatrix} \] \[ |A| = 2(0\cdot 4-1\cdot 1) - 1(1\cdot 4-1\cdot 4) + 2(1\cdot 1-0\cdot 4) \] \[ |A| = 2(-1)-1(0)+2(1) \] \[ |A|=-2+0+2=0 \] Therefore, \[ \operatorname{rank}(A)\lt 3 \]

Step 3: Check a non-zero minor of order \(2\).
Consider the minor formed by the first two rows and first two columns: \[ \begin{vmatrix} 2 & 1 \\ 1 & 0 \end{vmatrix} \] Its value is \[ 2\cdot 0-1\cdot 1=-1 \] Since \[ -1\neq 0, \] there exists a non-zero minor of order \(2\).
Therefore, \[ \operatorname{rank}(A)\geq 2 \]

Step 4: Determine the rank.
From Step 2, \[ \operatorname{rank}(A)\lt 3 \] From Step 3, \[ \operatorname{rank}(A)\geq 2 \] Hence, \[ \operatorname{rank}(A)=2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{2} \]