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: Multiply Matrices A and B
First, we need to calculate the product \(AB\).

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

To perform the multiplication, we will use the formula for matrix multiplication:

\[ AB_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj} \]

Calculating each element:

- For the element at position (1,1):
\[ AB_{11} = 3 \cdot 12 + (-5) \cdot 7 = 36 - 35 = 1 \]

- For the element at position (1,2):
\[ AB_{12} = 3 \cdot (-5) + (-5) \cdot (-3) = -15 + 15 = 0 \]

- For the element at position (2,1):
\[ AB_{21} = 7 \cdot 12 + (-12) \cdot 7 = 84 - 84 = 0 \]

- For the element at position (2,2):
\[ AB_{22} = 7 \cdot (-5) + (-12) \cdot (-3) = -35 + 36 = 1 \]

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

### Step 2: Find the Inverse of AB
Next, we need to find \((AB)^{-1}\). Since we have found that \(AB\) is the identity matrix \(I\):
\[ AB = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

The inverse of the identity matrix is itself:
\[ (AB)^{-1} = I^{-1} = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Final Answer
Thus, the value of \((AB)^{-1}\) is:
\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]