If the inverse of `[{:(1,p,q),(0,x,0),(0,0,1):}]"is"[{:(1,-p,-q),(0,1,0),(0,0,1):}]` then what is the value of x?

Options

A. 1
B. Zero
C. -1
D. `(1)/(p)+(1)/(q)`

Explanation

To find the value of \( x \) given that the inverse of matrix \( A \) is provided, we will follow these steps:

### Step 1: Define the matrices
Let matrix \( A \) be defined as:
\[ A = \begin{pmatrix} 1 & p & q \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
and its inverse \( A^{-1} \) is given by:
\[ A^{-1} = \begin{pmatrix} 1 & -p & -q \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Set up the equation
We know that the product of a matrix and its inverse equals the identity matrix:
\[ A \cdot A^{-1} = I \]
where \( I \) is the identity matrix:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Multiply the matrices
Now, we will multiply \( A \) and \( A^{-1} \):
\[ A \cdot A^{-1} = \begin{pmatrix} 1 & p & q \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -p & -q \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the elements of the resulting matrix:
- First row, first column: \( 1 \cdot 1 + p \cdot 0 + q \cdot 0 = 1 \)
- First row, second column: \( 1 \cdot (-p) + p \cdot 1 + q \cdot 0 = -p + p = 0 \)
- First row, third column: \( 1 \cdot (-q) + p \cdot 0 + q \cdot 1 = -q + q = 0 \)

- Second row, first column: \( 0 \cdot 1 + x \cdot 0 + 0 \cdot 0 = 0 \)
- Second row, second column: \( 0 \cdot (-p) + x \cdot 1 + 0 \cdot 0 = x \)
- Second row, third column: \( 0 \cdot (-q) + x \cdot 0 + 0 \cdot 1 = 0 \)

- Third row, first column: \( 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0 \)
- Third row, second column: \( 0 \cdot (-p) + 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Third row, third column: \( 0 \cdot (-q) + 0 \cdot 0 + 1 \cdot 1 = 1 \)

Thus, the product \( A \cdot A^{-1} \) results in:
\[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Set the resulting matrix equal to the identity matrix
We equate this result to the identity matrix:
\[ \begin{pmatrix} 1 & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Solve for \( x \)
From the second row, second column, we have:
\[ x = 1 \]

### Conclusion
Thus, the value of \( x \) is:
\[ \boxed{1} \]