If `A=[[-2, 0, 0], [-2, -1, 3]], B=[[0, 1], [2, 3], [1, -1]]`, then

Options

A. `(AB)^(-1)` not exist
B. `(AB)^(-1)` is null matrix
C. `(AB)^(-1)` exist
D. `(AB)^(-1)` unit matrix

Answer

To solve the problem, we need to find the product of matrices A and B, denoted as AB. Let's go through the steps systematically.

### Step 1: Define the matrices A and B
Given:
\[ A = \begin{bmatrix} -2 & 0 & 0 \\ -2 & -1 & 3 \end{bmatrix} \]
\[ B = \begin{bmatrix} 0 & 1 \\ 2 & 3 \\ 1 & -1 \end{bmatrix} \]

### Step 2: Check the dimensions of the matrices
Matrix A is of size 2x3 (2 rows and 3 columns) and matrix B is of size 3x2 (3 rows and 2 columns). Since the number of columns in A (3) is equal to the number of rows in B (3), we can multiply these matrices.

### Step 3: Multiply the matrices A and B
To find the product AB, we will use the formula for matrix multiplication. The element in the i-th row and j-th column of the resulting matrix is obtained by taking the dot product of the i-th row of A and the j-th column of B.

Calculating each element of the resulting matrix AB:

1. **Element (1,1)**:
\[ (-2 \cdot 0) + (0 \cdot 2) + (0 \cdot 1) = 0 + 0 + 0 = 0 \]

2. **Element (1,2)**:
\[ (-2 \cdot 1) + (0 \cdot 3) + (0 \cdot -1) = -2 + 0 + 0 = -2 \]

3. **Element (2,1)**:
\[ (-2 \cdot 0) + (-1 \cdot 2) + (3 \cdot 1) = 0 - 2 + 3 = 1 \]

4. **Element (2,2)**:
\[ (-2 \cdot 1) + (-1 \cdot 3) + (3 \cdot -1) = -2 - 3 - 3 = -8 \]

Thus, the product AB is:
\[ AB = \begin{bmatrix} 0 & -2 \\ 1 & -8 \end{bmatrix} \]

### Step 4: Determine if AB is a null matrix
A null matrix is a matrix where all elements are zero. Since \( AB \) contains non-zero elements, it is not a null matrix.

### Step 5: Calculate the determinant of AB
The determinant of a 2x2 matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is calculated as:
\[ \text{det}(AB) = ad - bc \]
For our matrix:
\[ \text{det}(AB) = (0 \cdot -8) - (-2 \cdot 1) = 0 + 2 = 2 \]

### Step 6: Determine if AB is a singular matrix
A matrix is singular if its determinant is zero. Since the determinant of \( AB \) is 2 (not zero), \( AB \) is a non-singular matrix.

### Step 7: Conclusion
Based on the calculations:
- \( AB \) is not a null matrix.
- The determinant of \( AB \) is 2 (not equal to 0), hence \( AB \) is non-singular.