Question

Let \(A = \begin{bmatrix} 2 & 1 & 4 3 & 2 & 5 0 & -1 & 1 \end{bmatrix}\) be a matrix, then the rank of \(A\) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

For any \(3 \times 3\) matrix:
- Calculate the determinant first. If it is non-zero, the rank is immediately 3.
- If the determinant is 0, the rank must be less than 3 (either 2, 1, or 0).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The rank of a matrix is the dimension of the vector space spanned by its columns or rows.
For a square matrix of order \(n \times n\), the rank is equal to \(n\) if and only if the matrix is non-singular (i.e., its determinant is non-zero).

Step 2: Key Formula or Approach:
We evaluate the determinant of the \(3 \times 3\) matrix \(A\).
If \(\det(A) \neq 0\), the rank of the matrix is 3.

Step 3: Detailed Explanation:
Let us write down the given matrix:
\[ A = \begin{bmatrix} 2 & 1 & 4 3 & 2 & 5 0 & -1 & 1 \end{bmatrix} \]
Now, calculate the determinant of \(A\) by expanding along the first row:
\[ \det(A) = 2 \left| \begin{matrix} 2 & 5 -1 & 1 \end{matrix} \right| - 1 \left| \begin{matrix} 3 & 5 0 & 1 \end{matrix} \right| + 4 \left| \begin{matrix} 3 & 2 0 & -1 \end{matrix} \right| \]
Evaluate each of the \(2 \times 2\) determinants:
\[ \left| \begin{matrix} 2 & 5 -1 & 1 \end{matrix} \right| = (2 \times 1) - (5 \times -1) = 2 - (-5) = 7 \]
\[ \left| \begin{matrix} 3 & 5 0 & 1 \end{matrix} \right| = (3 \times 1) - (5 \times 0) = 3 - 0 = 3 \]
\[ \left| \begin{matrix} 3 & 2 0 & -1 \end{matrix} \right| = (3 \times -1) - (2 \times 0) = -3 - 0 = -3 \]
Substitute these values back into the determinant expansion:
\[ \det(A) = 2(7) - 1(3) + 4(-3) \]
\[ \det(A) = 14 - 3 - 12 \]
\[ \det(A) = 14 - 15 = -1 \]
Since the determinant of \(A\) is \(-1\), which is not equal to zero:
\[ \det(A) \neq 0 \]
The matrix is non-singular, meaning all 3 row vectors (and column vectors) are linearly independent.
Therefore, the rank of the matrix is equal to its order, which is 3.

Step 4: Final Answer:
Therefore, the correct option is (D).