Suppose `|(x+1,3,5),(2,x+2,5),(2,3,x+4)|=0`, then the value of x is equal to

Options

A. `1,9`
B. `-1, 9`
C. `-1,-9`
D. `1,-9`

Answer

To solve the determinant equation \(|(x+1, 3, 5), (2, x+2, 5), (2, 3, x+4)| = 0\), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{vmatrix} \]

### Step 2: Apply Column Operations
We will simplify the determinant by adding all three columns to the first column:
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ D = \begin{vmatrix} (x+1) + 3 + 5 & 3 & 5 \\ 2 + (x+2) + 5 & x+2 & 5 \\ 2 + 3 + (x+4) & 3 & x+4 \end{vmatrix} \]
Calculating the first column:
- First row: \(x + 9\)
- Second row: \(x + 9\)
- Third row: \(x + 9\)

Thus, we have:
\[ D = \begin{vmatrix} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{vmatrix} \]

### Step 3: Factor Out Common Terms
We can factor out \(x + 9\) from the first column:
\[ D = (x + 9) \begin{vmatrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{vmatrix} \]

### Step 4: Simplify the 2x2 Determinant
Now we need to compute:
\[ \begin{vmatrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{vmatrix} \]
We can perform row operations to simplify this determinant. Subtract the first row from the second and third rows:
\[ D = (x + 9) \begin{vmatrix} 1 & 3 & 5 \\ 0 & (x+2 - 3) & 0 \\ 0 & (3 - 3) & (x+4 - 5) \end{vmatrix} \]
This simplifies to:
\[ D = (x + 9) \begin{vmatrix} 1 & 3 & 5 \\ 0 & x - 1 & 0 \\ 0 & 0 & x - 1 \end{vmatrix} \]

### Step 5: Calculate the Determinant
The determinant simplifies to:
\[ D = (x + 9)(x - 1)(x - 1) \]
Setting the determinant equal to zero:
\[ (x + 9)(x - 1)^2 = 0 \]

### Step 6: Solve for x
Setting each factor to zero gives us:
1. \(x + 9 = 0 \Rightarrow x = -9\)
2. \((x - 1)^2 = 0 \Rightarrow x = 1\)

### Final Answer
Thus, the values of \(x\) are:
\[ x = -9 \quad \text{or} \quad x = 1 \]