If `[(1,x,1)][(1,3,2),(0,5,1),(0,3,2)][(x),(1),(-2)]=[0]` then x is

Options

A. `-(1)/(2)`
B. `(1)/(2)`
C. 1
D. -1

Answer

To solve the equation given by the matrix multiplication, we follow these steps:

### Step 1: Set up the matrices
We have the following matrices:
- Row matrix: \( A = [1, x, 1] \)
- Square matrix:
\[ B = \begin{pmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{pmatrix} \]
- Column matrix:
\[ C = \begin{pmatrix} x \\ 1 \\ -2 \end{pmatrix} \]

We need to multiply these matrices and set the result equal to the zero matrix.

### Step 2: Multiply the square matrix \( B \) and the column matrix \( C \)
The multiplication of matrix \( B \) and matrix \( C \) is calculated as follows:
\[ B \cdot C = \begin{pmatrix} 1 & 3 & 2 \\ 0 & 5 & 1 \\ 0 & 3 & 2 \end{pmatrix} \cdot \begin{pmatrix} x \\ 1 \\ -2 \end{pmatrix} \]

Calculating each entry:
- First row:
\[ 1 \cdot x + 3 \cdot 1 + 2 \cdot (-2) = x + 3 - 4 = x - 1 \]
- Second row:
\[ 0 \cdot x + 5 \cdot 1 + 1 \cdot (-2) = 0 + 5 - 2 = 3 \]
- Third row:
\[ 0 \cdot x + 3 \cdot 1 + 2 \cdot (-2) = 0 + 3 - 4 = -1 \]

Thus, we have:
\[ B \cdot C = \begin{pmatrix} x - 1 \\ 3 \\ -1 \end{pmatrix} \]

### Step 3: Multiply the row matrix \( A \) with the result of \( B \cdot C \)
Now we multiply the row matrix \( A \) with the result:
\[ A \cdot (B \cdot C) = [1, x, 1] \cdot \begin{pmatrix} x - 1 \\ 3 \\ -1 \end{pmatrix} \]

Calculating this gives:
\[ 1 \cdot (x - 1) + x \cdot 3 + 1 \cdot (-1) = (x - 1) + 3x - 1 = 4x - 2 \]

### Step 4: Set the equation equal to zero
We set the result equal to zero:
\[ 4x - 2 = 0 \]

### Step 5: Solve for \( x \)
Solving the equation:
\[ 4x = 2 \\ x = \frac{2}{4} = \frac{1}{2} \]

Thus, the value of \( x \) is:
\[ \boxed{\frac{1}{2}} \]

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