find x, if
`[x" "-5" "-1][{:(1,0,2),(0,2,1),(2,0,3):}][{:(x),(4),(1):}]=0`

Explanation

To solve the equation given by the matrix multiplication, we need to find the value of \( x \) in the equation:

\[ [x, -5, -1] \cdot \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 \]

### Step 1: Matrix Multiplication
First, we will multiply the first matrix with the second matrix.

The first matrix is:
\[ A = \begin{bmatrix} x & -5 & -1 \end{bmatrix} \]

The second matrix is:
\[ B = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \]

We will calculate \( A \cdot B \):
\[ A \cdot B = \begin{bmatrix} x & -5 & -1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \]

Calculating the elements:
- First element: \( x \cdot 1 + (-5) \cdot 0 + (-1) \cdot 2 = x - 2 \)
- Second element: \( x \cdot 0 + (-5) \cdot 2 + (-1) \cdot 0 = -10 \)
- Third element: \( x \cdot 2 + (-5) \cdot 1 + (-1) \cdot 3 = 2x - 5 - 3 = 2x - 8 \)

Thus, we have:
\[ A \cdot B = \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \]

### Step 2: Multiply with the Third Matrix
Now we multiply the result with the third matrix:
\[ C = \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} \]

Calculating \( (A \cdot B) \cdot C \):
\[ \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \cdot \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} \]

Calculating the elements:
- First element: \( (x - 2)x + (-10) \cdot 4 + (2x - 8) \cdot 1 \)
\[ = x^2 - 2x - 40 + 2x - 8 = x^2 - 48 \]

### Step 3: Set the Equation to Zero
We set the result equal to zero:
\[ x^2 - 48 = 0 \]

### Step 4: Solve for x
Now we solve for \( x \):
\[ x^2 = 48 \]
\[ x = \pm \sqrt{48} = \pm 4\sqrt{3} \]

### Final Answer
Thus, the values of \( x \) are:
\[ x = 4\sqrt{3} \quad \text{or} \quad x = -4\sqrt{3} \]

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