If `A=[(1,-1,0),(2,3,4),(0,1,2)] and B=[(2,2,-4),(-4,2,-4),(2,-1,5)]` , then :

Options

A. `A^(-1) = B`
B. `A^(-1)= 6B`
C. `B^(-1) = B`
D. `B^(-1)=1/6A`

Answer

To solve the problem, we need to find the relationship between the matrices A and B given the matrices:

\[ A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]
\[ B = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]

We will first compute the product of matrices A and B (i.e., \( AB \)) and then analyze the result to determine which of the given options is correct.

### Step 1: Calculate the product \( AB \)

To find \( AB \), we perform matrix multiplication:

1. **First Row of A with Columns of B**:
- For the first element of \( AB \):
\[ 1 \cdot 2 + (-1) \cdot (-4) + 0 \cdot 2 = 2 + 4 + 0 = 6 \]
- For the second element of \( AB \):
\[ 1 \cdot 2 + (-1) \cdot 2 + 0 \cdot (-1) = 2 - 2 + 0 = 0 \]
- For the third element of \( AB \):
\[ 1 \cdot (-4) + (-1) \cdot (-4) + 0 \cdot 5 = -4 + 4 + 0 = 0 \]

2. **Second Row of A with Columns of B**:
- For the first element of \( AB \):
\[ 2 \cdot 2 + 3 \cdot (-4) + 4 \cdot 2 = 4 - 12 + 8 = 0 \]
- For the second element of \( AB \):
\[ 2 \cdot 2 + 3 \cdot 2 + 4 \cdot (-1) = 4 + 6 - 4 = 6 \]
- For the third element of \( AB \):
\[ 2 \cdot (-4) + 3 \cdot (-4) + 4 \cdot 5 = -8 - 12 + 20 = 0 \]

3. **Third Row of A with Columns of B**:
- For the first element of \( AB \):
\[ 0 \cdot 2 + 1 \cdot (-4) + 2 \cdot 2 = 0 - 4 + 4 = 0 \]
- For the second element of \( AB \):
\[ 0 \cdot 2 + 1 \cdot 2 + 2 \cdot (-1) = 0 + 2 - 2 = 0 \]
- For the third element of \( AB \):
\[ 0 \cdot (-4) + 1 \cdot (-4) + 2 \cdot 5 = 0 - 4 + 10 = 6 \]

Putting it all together, we find:

\[ AB = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I \]

### Step 2: Relate \( A \) and \( B \)

From the equation \( AB = 6I \), we can multiply both sides by \( B^{-1} \):

\[ AB B^{-1} = 6I B^{-1} \]

This simplifies to:

\[ A = 6I B^{-1} \]

Rearranging gives us:

\[ B^{-1} = \frac{1}{6} A \]

### Conclusion

Thus, we find that:

\[ B^{-1} = \frac{1}{6} A \]

This corresponds to option 4: \( B^{-1} = \frac{1}{6} B \).