If `|{:(x+1,1,1),(1,x+1,1),(-1,1,x+1):}|=0`, find the value of 'x'.

Answer

To solve the determinant equation \( | \begin{pmatrix} x+1 & 1 & 1 \\ 1 & x+1 & 1 \\ -1 & 1 & x+1 \end{pmatrix} | = 0 \), we will follow these steps:

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

### Step 2: Apply Column Operations
We can simplify the determinant by performing column operations. Let's replace \( C_2 \) with \( C_2 - C_3 \):
\[ D = \begin{vmatrix} x+1 & 1-1 & 1 \\ 1 & x+1-1 & 1 \\ -1 & 1-1 & x+1 \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} x+1 & 0 & 1 \\ 1 & x & 1 \\ -1 & 0 & x+1 \end{vmatrix} \]

### Step 3: Apply Row Operations
Next, we can perform a row operation by replacing \( R_3 \) with \( R_3 + R_2 \):
\[ D = \begin{vmatrix} x+1 & 0 & 1 \\ 1 & x & 1 \\ 0 & x & x+2 \end{vmatrix} \]

### Step 4: Expand the Determinant
Now, we can expand the determinant along the first column:
\[ D = (x+1) \begin{vmatrix} x & 1 \\ x & x+2 \end{vmatrix} - 0 + 0 \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} x & 1 \\ x & x+2 \end{vmatrix} = x(x+2) - 1(x) = x^2 + 2x - x = x^2 + x \]
Thus,
\[ D = (x+1)(x^2 + x) \]

### Step 5: Set the Determinant to Zero
Now, we set the determinant equal to zero:
\[ (x+1)(x^2 + x) = 0 \]

### Step 6: Solve for x
This gives us two factors:
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x^2 + x = 0 \) → \( x(x + 1) = 0 \) → \( x = 0 \) or \( x = -1 \)

Thus, the solutions are:
- \( x = 0 \)
- \( x = -1 \)
- \( x = -2 \) (from the factorization of \( x^2 + x = 0 \))

### Final Values of x
The values of \( x \) that satisfy the determinant being zero are:
\[ x = 0, -1, -2 \]