Question

The rank of the matrix \[ A= \begin{pmatrix} 1&2&3\\ 2&1&0\\ 0&1&2 \end{pmatrix} \]is

• A. 0
• B. 1
• C. 3
• D. 2

Hint:

For a \(3\times3\) matrix: \[ \boxed{ \begin{aligned} \det(A)\neq0 &\Rightarrow \text{Rank}=3 \det(A)=0 &\Rightarrow \text{Check row reduction} \end{aligned} } \] Here, \[ R_2=-3R_3, \] so only two independent rows exist.

Answer 1

The Correct Option is D

Concept:

The rank of a matrix is the maximum number of linearly independent rows (or columns).

For a \(3 \times 3\) matrix:

• If \(\det(A) \neq 0\), then \(\operatorname{Rank}(A)=3\).
• If \(\det(A)=0\), reduce the matrix to echelon form.

Step 1: Find the determinant.

\[ \begin{aligned} |A| &=1 \begin{vmatrix} 1&0\\ 1&2 \end{vmatrix} -2 \begin{vmatrix} 2&0\\ 0&2 \end{vmatrix} +3 \begin{vmatrix} 2&1\\ 0&1 \end{vmatrix} \\ &=1(2)-2(4)+3(2) \\ &=2-8+6 \\ &=0. \end{aligned} \]

Hence, \(\det(A)=0\), so the rank is less than 3.

Step 2: Reduce the matrix.

Apply:

\[ R_2 \rightarrow R_2 - 2R_1 \] \[ \begin{pmatrix} 1&2&3\\ 0&-3&-6\\ 0&1&2 \end{pmatrix} \]

Now,

\[ R_2=-3R_3. \]

Thus, only two rows are linearly independent.

Therefore,

\[ \operatorname{Rank}(A)=2. \]

Hence, the correct option is (D).