Let A be an `mxxn` matrix. If there exists a matrix L of type `nxxm` such that `LA=I_(n)`, then L is called left inverse of A. Which of the following matrices is NOT left inverse of matrix `[(1,-1),(1,1),(2,3)]?`

Options

A. `[(1/2,1/2,0),(-1/2,1/2,0)]`
B. `[(2,-7,3),(-1/2,1/2,0)]`
C. `[(-1/2,1/2,0),(-1/2,1/2,0)]`
D. `[(0,3,-1),(-1/2,1/2,0)]`

Explanation

To determine which of the given matrices is NOT a left inverse of the matrix \( A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 2 & 3 \end{pmatrix} \), we need to check each option to see if it satisfies the condition \( LA = I_n \), where \( I_n \) is the identity matrix of order \( n \).

### Step-by-Step Solution:

1. **Identify the Matrix A**:
\[ A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 2 & 3 \end{pmatrix} \]
This is a \( 3 \times 2 \) matrix.

2. **Identify the Identity Matrix \( I_n \)**:
Since \( A \) is \( 3 \times 2 \), the left inverse \( L \) must be \( 2 \times 3 \) and the identity matrix \( I_n \) will be \( 2 \times 2 \):
\[ I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

3. **Check Each Option**:
We will multiply each option \( L \) by \( A \) and check if the result is \( I_2 \).

**Option A**:
\[ L_A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix} \]
\[ L_A \cdot A = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 2 & 3 \end{pmatrix} \]
Calculate the product:
- First row: \( \frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 1 \)
- Second row: \( -\frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 0 \)
- Result: \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 \)

**Option B**:
\[ L_B = \begin{pmatrix} 2 & -7 & 3 \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix} \]
Calculate \( L_B \cdot A \):
- First row: \( 2(1) + (-7)(1) + 3(2) = 2 - 7 + 6 = 1 \)
- Second row: \( -\frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 0 \)
- Result: \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 \)

**Option C**:
\[ L_C = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix} \]
Calculate \( L_C \cdot A \):
- First row: \( -\frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 0 \)
- Second row: \( -\frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 0 \)
- Result: \( \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \neq I_2 \)

**Option D**:
\[ L_D = \begin{pmatrix} 0 & 3 & -1 \\ -\frac{1}{2} & \frac{1}{2} & 0 \end{pmatrix} \]
Calculate \( L_D \cdot A \):
- First row: \( 0(1) + 3(1) + (-1)(2) = 3 - 2 = 1 \)
- Second row: \( -\frac{1}{2}(1) + \frac{1}{2}(1) + 0(2) = 0 \)
- Result: \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I_2 \)

4. **Conclusion**:
The matrix that is NOT a left inverse of \( A \) is **Option C**.