If `A = [[a,b,c],[x,y,z],[p,q,r]], B= [[q , -b,y],[-p,a,-x],[r,-c,z]]` and if A is
invertible, then which of the following is not true?

Options

A. `abs(A) = abs(B)`
B. `abs(A) = -abs(B)`
C. `abs(adjA) = abs(adjB)`
D. A is invertible `hArr` B is invertble

Explanation

To solve the problem, we need to analyze the matrices \( A \) and \( B \) given in the question and understand the implications of the invertibility of matrix \( A \).

### Step-by-Step Solution:

1. **Define the Matrices**:
\[ A = \begin{bmatrix} a & b & c \\ x & y & z \\ p & q & r \end{bmatrix}, \quad B = \begin{bmatrix} q & -b & y \\ -p & a & -x \\ r & -c & z \end{bmatrix} \]

2. **Understanding Invertibility**:
Since \( A \) is given to be invertible, this implies that the determinant of \( A \) is non-zero:
\[ \text{det}(A) \neq 0 \]

3. **Calculate the Determinant of Matrix \( B \)**:
We can express the determinant of \( B \) in terms of the determinant of \( A \). To do this, we can perform row and column operations on \( B \) to relate it to \( A \).

4. **Row Operations**:
- Take \(-1\) common from the second row of \( B \):
\[ \text{det}(B) = -\text{det}\begin{bmatrix} q & -b & y \\ p & -a & x \\ r & -c & z \end{bmatrix} \]

5. **Column Operations**:
- Take \(-1\) common from the second column of \( B \):
\[ \text{det}(B) = (-1)(-1)\text{det}\begin{bmatrix} q & b & y \\ p & a & x \\ r & c & z \end{bmatrix} = \text{det}\begin{bmatrix} q & b & y \\ p & a & x \\ r & c & z \end{bmatrix} \]

6. **Relate \( B \) to \( A \)**:
By rearranging the columns and rows, we can show that:
\[ \text{det}(B) = -\text{det}(A) \]

7. **Conclusion on Invertibility**:
Since \( \text{det}(A) \neq 0 \), it follows that \( \text{det}(B) \neq 0 \) as well. Therefore, both matrices \( A \) and \( B \) are invertible.

8. **Evaluate the Statements**:
We need to determine which of the following statements is not true:
- \( \text{det}(A) = \text{det}(B) \)
- \( \text{det}(A) = -\text{det}(B) \)
- \( \text{adj}(A) = \text{adj}(B) \)
- If \( A \) is invertible, then \( B \) is invertible.
- If \( B \) is invertible, then \( A \) is invertible.

From our analysis, we found that:
- \( \text{det}(A) = -\text{det}(B) \) is true.
- \( \text{det}(A) = \text{det}(B) \) is not true.

### Final Answer:
The statement that is **not true** is:
\[ \text{det}(A) = \text{det}(B) \]