Find AB, if `A=[{:(1,2,3),(1,-2,3):}]` and `B=[{:(1,-1),(1,2),(1,-2):}]`. Examine whether `(AB)^(-1)` exist or not.

Answer

To solve the problem of finding the product of matrices \( A \) and \( B \), and then determining whether the inverse of the product \( AB \) exists, we will follow these steps:

### Step 1: Define the matrices \( A \) and \( B \)

Given:
\[ A = \begin{pmatrix} 1 & 2 & 3 \\ 1 & -2 & 3 \end{pmatrix} \]
\[ B = \begin{pmatrix} 1 & -1 \\ 1 & 2 \\ 1 & -2 \end{pmatrix} \]

### Step 2: Check the dimensions of the matrices

Matrix \( A \) is a \( 2 \times 3 \) matrix (2 rows and 3 columns), and matrix \( B \) is a \( 3 \times 2 \) matrix (3 rows and 2 columns). The product \( AB \) can be computed since the number of columns in \( A \) (3) matches the number of rows in \( B \) (3).

### Step 3: Calculate the product \( AB \)

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

\[ AB = \begin{pmatrix} 1 & 2 & 3 \\ 1 & -2 & 3 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 2 \\ 1 & -2 \end{pmatrix} \]

Calculating each element of the resulting \( 2 \times 2 \) matrix:

- First row, first column:
\[ 1 \cdot 1 + 2 \cdot 1 + 3 \cdot 1 = 1 + 2 + 3 = 6 \]

- First row, second column:
\[ 1 \cdot (-1) + 2 \cdot 2 + 3 \cdot (-2) = -1 + 4 - 6 = -3 \]

- Second row, first column:
\[ 1 \cdot 1 + (-2) \cdot 1 + 3 \cdot 1 = 1 - 2 + 3 = 2 \]

- Second row, second column:
\[ 1 \cdot (-1) + (-2) \cdot 2 + 3 \cdot (-2) = -1 - 4 - 6 = -11 \]

Thus, the product \( AB \) is:
\[ AB = \begin{pmatrix} 6 & -3 \\ 2 & -11 \end{pmatrix} \]

### Step 4: Determine if \( (AB)^{-1} \) exists

To check if the inverse of \( AB \) exists, we need to calculate the determinant of \( AB \).

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{det}(AB) = ad - bc \]

For our matrix \( AB \):
\[ a = 6, \quad b = -3, \quad c = 2, \quad d = -11 \]
Calculating the determinant:
\[ \text{det}(AB) = (6)(-11) - (-3)(2) = -66 + 6 = -60 \]

Since the determinant \( \text{det}(AB) = -60 \) is not equal to zero, the inverse \( (AB)^{-1} \) exists.

### Final Result

The product \( AB \) is:
\[ AB = \begin{pmatrix} 6 & -3 \\ 2 & -11 \end{pmatrix} \]
And the inverse \( (AB)^{-1} \) exists.

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