If $\begin{vmatrix}1+x&1&1\ 1&1+x&1\ 1&1&1+x\end{vmatrix}=0$ then $x=$

Question

Accepted Answer

To solve the problem, we need to evaluate the determinant of the given matrix and find the values of $ x $ that make the determinant equal to zero. The matrix is:

$$
\begin{vmatrix}
1+x & 1 & 1 \
1 & 1+x & 1 \
1 & 1 & 1+x
\end{vmatrix}
$$

The determinant of a 3x3 matrix $\begin{vmatrix} a & b & c \ d & e & f \ g & h & i \end{vmatrix}$ is calculated as:

$$
a(ei - fh) - b(di - fg) + c(dh - eg)
$$

Applying this formula to our matrix, we have:

$$
(1+x)((1+x)(1+x) - 1 \cdot 1) - 1(1(1+x) - 1 \cdot 1) + 1(1 \cdot 1 - (1+x) \cdot 1)
$$

Simplifying each term:

1. $(1+x)((1+x)^2 - 1)$
2. $-1((1+x) - 1)$
3. $1(1 - (1+x))$

Let's simplify each part:

1. $(1+x)((1+x)^2 - 1) = (1+x)(1 + 2x + x^2 - 1) = (1+x)(x^2 + 2x) = (1+x)x(x+2)$
2. $-1((1+x) - 1) = -1(x)$
3. $1(1 - (1+x)) = 1(-x) = -x$

Combine these:

$$
(1+x)x(x+2) - x - x = (1+x)x(x+2) - 2x
$$

Set the determinant to zero:

$$
(1+x)x(x+2) - 2x = 0
$$

Factor out $x$:

$$
x((1+x)(x+2) - 2) = 0
$$

This gives us two cases:

1. $x = 0$
2. $(1+x)(x+2) - 2 = 0$

Solve the second equation:

$$
(1+x)(x+2) - 2 = x^2 + 2x + x + 2 - 2 = x^2 + 3x = 0
$$

Factor:

$$
x(x+3) = 0
$$

This gives solutions $x = 0$ and $x = -3$.

Thus, the values of $x$ that satisfy the determinant being zero are $x = 0$ and $x = -3$. Therefore, the correct answer is Option 3: 0, -3.

**Explanation for Incorrect Options:**

- **Option 1 (0, 1, 3):** This option includes 1 and 3, which do not satisfy the determinant equation.
- **Option 2 (0, 1, -3):** This option includes 1, which does not satisfy the determinant equation.
- **Option 4 (0, 3):** This option includes 3, which does not satisfy the determinant equation.

Thus, Option 3 is the only correct choice, as it includes all and only the values that make the determinant zero.