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. Similarly, if there exists a matrix R of type `nxxm` such that `AR=I_(m)`, then R is called right inverse of A.
For example, to find right inverse of matrix
`A=[(1,-1),(1,1),(2,3)]`, we take `R=[(x,y,x),(u,v,w)]`
and solve`AR=I_(3)`, i.e.,
`[(1,-1),(1,1),(2,3)][(x,y,z),(u,v,w)]=[(1,0,0),(0,1,0),(0,0,1)]`
`{:(implies,x-u=1,y-v=0,z-w=0),(,x+u=0,y+v=1,z+w=0),(,2x+3u=0,2y+3v=0,2z+3w=1):}`
As this system of equations is inconsistent, we say there is no right inverse for matrix A.
For which of the following matrices, the number of left inverses is greater than the number of right inverses ?

Options

A. `[(1,2,4),(-3,2,1)]`
B. `[(3,2,1),(3,2,1)]`
C. `[(1,4),(2,-3),(2,-3)]`
D. `[(3,3),(1,1),(4,4)]`

Explanation

To solve the problem, we need to determine for which of the given matrices the number of left inverses is greater than the number of right inverses. Let's analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- A matrix \( A \) has a left inverse \( L \) if \( LA = I_n \).
- A matrix \( A \) has a right inverse \( R \) if \( AR = I_m \).

2. **Analyzing the Options**:
- We have four matrices to consider:
- **Option A**:
\[ A = \begin{pmatrix} 1 & 2 \\ 4 & -3 \\ 2 & 1 \end{pmatrix} \]
- **Option B**:
\[ B = \begin{pmatrix} 3 & 2 \\ 1 & 3 \\ 2 & 1 \end{pmatrix} \]
- **Option C**:
\[ C = \begin{pmatrix} 1 & 4 \\ -3 & 2 \\ -3 & 1 \end{pmatrix} \]
- **Option D**:
\[ D = \begin{pmatrix} 3 & 3 \\ 1 & 1 \\ 4 & 4 \end{pmatrix} \]

3. **Checking for Left Inverses**:
- For **Option B** and **Option D**, we observe that the rows are linearly dependent (the rows of B are the same, and the rows of D are multiples of each other). Therefore, these matrices do not have a left inverse.
- For **Option A** and **Option C**, we need to check if left inverses exist.

4. **Finding Left Inverse for Option A**:
- Let \( L = \begin{pmatrix} A & B \\ C & D \\ E & F \end{pmatrix} \).
- We need to solve \( LA = I_3 \):
\[ \begin{pmatrix} A & B \\ C & D \\ E & F \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 4 & -3 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
- This leads to a system of equations which we find to be inconsistent.

5. **Finding Left Inverse for Option C**:
- Let \( L = \begin{pmatrix} A & B \\ C & D \\ E & F \end{pmatrix} \).
- We need to solve \( LA = I_3 \):
\[ \begin{pmatrix} A & B \\ C & D \\ E & F \end{pmatrix} \begin{pmatrix} 1 & 4 \\ -3 & 2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
- This leads to a system of equations that can be solved, yielding infinite solutions.

6. **Finding Right Inverses**:
- For **Option A**, we find that there is no right inverse.
- For **Option C**, we also find that there is no right inverse.

### Conclusion:
- **Option A** has no left inverse and no right inverse.
- **Option B** has no left or right inverse.
- **Option C** has infinite left inverses and no right inverse.
- **Option D** has no left or right inverse.

Thus, the option where the number of left inverses is greater than the number of right inverses is **Option C**.