Find the value of `x` such that `[[1, x, 1]][[1 ,3, 2],[2, 5 ,1],[ 15 ,3 ,2]][[1], [2],[x]]=0`

Explanation

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

### Step 1: Define the matrices
Let:
- \( A = \begin{bmatrix} 1 & x & 1 \end{bmatrix} \) (1x3 matrix)
- \( B = \begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{bmatrix} \) (3x3 matrix)
- \( C = \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} \) (3x1 matrix)

### Step 2: Multiply matrices A and B
To find the product \( AB \):
\[ AB = \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 15 \end{bmatrix} = 1 \cdot 1 + x \cdot 2 + 1 \cdot 15 = 1 + 2x + 15 = 16 + 2x \]
\[ AB = \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \\ 3 \end{bmatrix} = 1 \cdot 3 + x \cdot 5 + 1 \cdot 3 = 3 + 5x + 3 = 6 + 5x \]
\[ AB = \begin{bmatrix} 1 & x & 1 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix} = 1 \cdot 2 + x \cdot 1 + 1 \cdot 2 = 2 + x + 2 = 4 + x \]

Thus, we have:
\[ AB = \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \]

### Step 3: Multiply the result with matrix C
Now, we multiply \( AB \) with \( C \):
\[ (AB)C = \begin{bmatrix} 16 + 2x & 6 + 5x & 4 + x \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ x \end{bmatrix} \]
Calculating this gives:
\[ = (16 + 2x) \cdot 1 + (6 + 5x) \cdot 2 + (4 + x) \cdot x \]
\[ = 16 + 2x + 12 + 10x + 4 + x^2 \]
\[ = x^2 + 13x + 28 \]

### Step 4: Set the equation to zero
We need this product to equal the null matrix (0):
\[ x^2 + 13x + 28 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 13, c = 28 \):
\[ x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 1 \cdot 28}}{2 \cdot 1} \]
\[ = \frac{-13 \pm \sqrt{169 - 112}}{2} \]
\[ = \frac{-13 \pm \sqrt{57}}{2} \]

### Step 6: Calculate the roots
The roots are:
\[ x = \frac{-13 + \sqrt{57}}{2} \quad \text{and} \quad x = \frac{-13 - \sqrt{57}}{2} \]

### Final Result
Thus, the values of \( x \) are:
\[ x = \frac{-13 + \sqrt{57}}{2} \quad \text{and} \quad x = \frac{-13 - \sqrt{57}}{2} \]

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