Consider two matrices `A=[{:(1,2),(2,1),(1,1):}]and B=[{:(1,2,-4),(2,1,-4):}]`. Which one of the following is correct ?

Options

A. B is the right inverse of A
B. B is the left inverse of A
C. B is the both sided inverse of A
D. None of the above

Explanation

To solve the problem, we need to determine the relationship between the matrices A and B, specifically whether B is a left inverse, right inverse, or both for the matrix A.

### Step-by-Step Solution:

1. **Identify the matrices A and B:**
- Matrix A is given as:
\[ A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 1 \end{pmatrix} \]
- Matrix B is given as:
\[ B = \begin{pmatrix} 1 & 2 & -4 \\ 2 & 1 & -4 \end{pmatrix} \]

2. **Determine the dimensions of A and B:**
- Matrix A is a \(3 \times 2\) matrix (3 rows and 2 columns).
- Matrix B is a \(2 \times 3\) matrix (2 rows and 3 columns).

3. **Calculate the product \(AB\):**
- The product \(AB\) will be a \(3 \times 3\) matrix since the number of columns in A (2) matches the number of rows in B (2).
- Calculate \(AB\):
\[ AB = \begin{pmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & -4 \\ 2 & 1 & -4 \end{pmatrix} \]
- Performing the multiplication:
- First row, first column: \(1 \cdot 1 + 2 \cdot 2 = 1 + 4 = 5\)
- First row, second column: \(1 \cdot 2 + 2 \cdot 1 = 2 + 2 = 4\)
- First row, third column: \(1 \cdot -4 + 2 \cdot -4 = -4 - 8 = -12\)
- Second row, first column: \(2 \cdot 1 + 1 \cdot 2 = 2 + 2 = 4\)
- Second row, second column: \(2 \cdot 2 + 1 \cdot 1 = 4 + 1 = 5\)
- Second row, third column: \(2 \cdot -4 + 1 \cdot -4 = -8 - 4 = -12\)
- Third row, first column: \(1 \cdot 1 + 1 \cdot 2 = 1 + 2 = 3\)
- Third row, second column: \(1 \cdot 2 + 1 \cdot 1 = 2 + 1 = 3\)
- Third row, third column: \(1 \cdot -4 + 1 \cdot -4 = -4 - 4 = -8\)
- Thus, we have:
\[ AB = \begin{pmatrix} 5 & 4 & -12 \\ 4 & 5 & -12 \\ 3 & 3 & -8 \end{pmatrix} \]

4. **Calculate the product \(BA\):**
- The product \(BA\) will be a \(2 \times 2\) matrix since the number of columns in B (3) matches the number of rows in A (3).
- Calculate \(BA\):
\[ BA = \begin{pmatrix} 1 & 2 & -4 \\ 2 & 1 & -4 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & 1 \\ 1 & 1 \end{pmatrix} \]
- Performing the multiplication:
- First row, first column: \(1 \cdot 1 + 2 \cdot 2 + -4 \cdot 1 = 1 + 4 - 4 = 1\)
- First row, second column: \(1 \cdot 2 + 2 \cdot 1 + -4 \cdot 1 = 2 + 2 - 4 = 0\)
- Second row, first column: \(2 \cdot 1 + 1 \cdot 2 + -4 \cdot 1 = 2 + 2 - 4 = 0\)
- Second row, second column: \(2 \cdot 2 + 1 \cdot 1 + -4 \cdot 1 = 4 + 1 - 4 = 1\)
- Thus, we have:
\[ BA = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]
- This is the identity matrix of order 2.

5. **Conclusion:**
- Since \(BA = I\) (identity matrix), B is a left inverse of A.
- Therefore, the correct answer is that B is the left inverse of A.

### Final Answer:
B is the left inverse of A.