Question

Let $ X = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \middle| a, b, c, d \in \mathbb{R} \right\} $. If $ f: X \to \mathbb{R} $ is defined by $ f(A) = \det(A) $ for all $ A \in X $, then $ f $ is

• A. one-one but not onto
• B. onto but not one-one
• C. one-one and onto
• D. neither one-one nor onto

Answer 1

The Correct Option is B

To solve the problem, we need to analyze the function \( f : X \rightarrow \mathbb{R} \) defined by \( f(A) = \det(A) \), where \( A \in X \) and \( X \) is the set of all 2x2 matrices of the form:

\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \quad \text{with } a, b, c, d \in \mathbb{R} \]

1. Understanding the Function:
The function maps each 2x2 real matrix to a real number which is the determinant of the matrix:
\[ \det(A) = ad - bc \]

2. Determine If the Function Is Onto:
We want to see if for every real number \( r \), there exists a matrix \( A \in X \) such that \( \det(A) = r \).
Take a matrix \( A = \begin{bmatrix} 1 & 0 \\ 0 & r \end{bmatrix} \). Then, \( \det(A) = 1 \cdot r - 0 \cdot 0 = r \).
Hence, any real number \( r \) can be obtained as the determinant of some matrix. So, \( f \) is onto.

3. Determine If the Function Is One-One:
If \( f \) were one-one, then different matrices would have different determinants.
But consider matrices:
\( A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( A_2 = \begin{bmatrix} 2 & 0 \\ 0 & 0.5 \end{bmatrix} \).
Both have \( \det(A_1) = \det(A_2) = 1 \), but \( A_1 \neq A_2 \).
So, \( f \) is not one-one.

Final Answer:
The function is onto but not one-one.