Question

Let \( M \) be the set of all \( 2 \times 2 \) matrices with entries from the set \( R \) of real numbers. Then the function \( f : M \to R \) defined by \( f(A) = |A| \) for every \( A \in M \) is

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

Hint:

Determinant maps matrices to real numbers, but different matrices can have the same determinant, so injectivity usually fails.

Answer 1

The Correct Option is C

Step 1: Understand the function.
The function is defined as:
\[ f(A) = |A|, \]
where \( |A| \) denotes the determinant of a \(2 \times 2\) matrix.
If:
\[ A = \begin{pmatrix} a & b c & d \end{pmatrix}, \] then:
\[ |A| = ad - bc. \]

Step 2: Check whether the function is one-one.
A function is one-one if different inputs give different outputs.
Consider two different matrices:
\[ A = \begin{pmatrix} 1 & 0 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 0 0 & \frac{1}{2} \end{pmatrix}. \]
Now:
\[ |A| = 1, \quad |B| = 1. \]
Thus, two different matrices have the same determinant.
Hence, the function is not one-one.

Step 3: Check whether the function is onto.
A function is onto if every real number is an image of some matrix.

Step 4: Construct a matrix for any real number.
Let \( k \in R \). Consider the matrix:
\[ A = \begin{pmatrix} k & 0 0 & 1 \end{pmatrix}. \]

Step 5: Compute its determinant.
\[ |A| = k \cdot 1 - 0 = k. \]

Step 6: Interpret the result.
Since for every real number \( k \), there exists a matrix whose determinant is \( k \), the function covers all real numbers.
Thus, the function is onto.

Step 7: Final conclusion.
The function is onto but not one-one.
Final Answer:
\[ \boxed{\text{onto but not one-one}}. \]