Let `A=[(3,-5),(7,-12)] and B=[(12,-5),(7,-3)]` bc two given matrices, then `(AB)^(-1)` is :

Options

A. `[(1,0),(0,-1)]`
B. `[(1,0),(0,1)]`
C. `[(-1,0),(0,1)]`
D. `[(0,1),(1,0)]`

Answer

To solve the problem of finding \((AB)^{-1}\) where \(A = \begin{pmatrix} 3 & -5 \\ 7 & -12 \end{pmatrix}\) and \(B = \begin{pmatrix} 12 & -5 \\ 7 & -3 \end{pmatrix}\), we will follow these steps:

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

To find the product of two matrices \(A\) and \(B\), we use the formula for matrix multiplication. The element at position \((i, j)\) in the resulting matrix is calculated by taking the dot product of the \(i^{th}\) row of the first matrix and the \(j^{th}\) column of the second matrix.

\[ AB = \begin{pmatrix} 3 & -5 \\ 7 & -12 \end{pmatrix} \begin{pmatrix} 12 & -5 \\ 7 & -3 \end{pmatrix} \]

Calculating the elements of the product:

1. First row, first column:
\[ 3 \cdot 12 + (-5) \cdot 7 = 36 - 35 = 1 \]

2. First row, second column:
\[ 3 \cdot (-5) + (-5) \cdot (-3) = -15 + 15 = 0 \]

3. Second row, first column:
\[ 7 \cdot 12 + (-12) \cdot 7 = 84 - 84 = 0 \]

4. Second row, second column:
\[ 7 \cdot (-5) + (-12) \cdot (-3) = -35 + 36 = 1 \]

Thus, the product \(AB\) is:
\[ AB = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Find \((AB)^{-1}\)

The inverse of a matrix is defined such that:
\[ A \cdot A^{-1} = I \]
where \(I\) is the identity matrix. Since we have already calculated \(AB\) and found that:
\[ AB = I \]
it follows that:
\[ (AB)^{-1} = I \]
Thus,
\[ (AB)^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Conclusion

The answer is:
\[ (AB)^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Options

Looking at the options provided, we find that this corresponds to:
- Option B: \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)