The equation `|{:(x-2,3,1),(4x-2,10,4),(2x-1,5,1):}|=0` is satisfied by

Options

A. `x=-2`
B. `x=-5`
C. `x=-7`
D. `x=-9`

Answer

To solve the equation given by the determinant \( | \begin{pmatrix} x-2 & 3 & 1 \\ 4x-2 & 10 & 4 \\ 2x-1 & 5 & 1 \end{pmatrix} | = 0 \), we will follow these steps:

### Step 1: Expand the Determinant
We will expand the determinant along the first row.

\[ |A| = (x-2) \cdot | \begin{pmatrix} 10 & 4 \\ 5 & 1 \end{pmatrix} | - 3 \cdot | \begin{pmatrix} 4x-2 & 4 \\ 2x-1 & 1 \end{pmatrix} | + 1 \cdot | \begin{pmatrix} 4x-2 & 10 \\ 2x-1 & 5 \end{pmatrix} | \]

### Step 2: Calculate the 2x2 Determinants
Now, we will calculate the 2x2 determinants.

1. For \( | \begin{pmatrix} 10 & 4 \\ 5 & 1 \end{pmatrix} | \):
\[ = (10 \cdot 1) - (4 \cdot 5) = 10 - 20 = -10 \]

2. For \( | \begin{pmatrix} 4x-2 & 4 \\ 2x-1 & 1 \end{pmatrix} | \):
\[ = ((4x-2) \cdot 1) - (4 \cdot (2x-1)) = (4x - 2) - (8x - 4) = 4x - 2 - 8x + 4 = -4x + 2 \]

3. For \( | \begin{pmatrix} 4x-2 & 10 \\ 2x-1 & 5 \end{pmatrix} | \):
\[ = ((4x-2) \cdot 5) - (10 \cdot (2x-1)) = (20x - 10) - (20x - 10) = 0 \]

### Step 3: Substitute Back into the Determinant
Now substitute these values back into the determinant expansion:

\[ |A| = (x-2)(-10) - 3(-4x + 2) + 1(0) \]
\[ = -10(x-2) + 12x - 6 \]
\[ = -10x + 20 + 12x - 6 \]
\[ = 2x + 14 \]

### Step 4: Set the Determinant to Zero
Now we set the determinant equal to zero:

\[ 2x + 14 = 0 \]

### Step 5: Solve for x
Now, solve for \( x \):

\[ 2x = -14 \]
\[ x = -7 \]

### Final Answer
The equation \( | \begin{pmatrix} x-2 & 3 & 1 \\ 4x-2 & 10 & 4 \\ 2x-1 & 5 & 1 \end{pmatrix} | = 0 \) is satisfied by \( x = -7 \).

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