Find the value of x for which the matrix `A=[(2//x,-1,2),(1,x,2x^(2)),(1,1//x,2)]` is singular.

Explanation

To find the value of \( x \) for which the matrix

\[ A = \begin{pmatrix} \frac{2}{x} & -1 & 2 \\ 1 & x & 2x^2 \\ 1 & \frac{1}{x} & 2 \end{pmatrix} \]

is singular, we need to compute the determinant of the matrix \( A \) and set it equal to zero. A matrix is singular if its determinant is zero.

### Step 1: Calculate the Determinant

The determinant of a \( 3 \times 3 \) matrix can be calculated using the formula:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = \frac{2}{x} \), \( b = -1 \), \( c = 2 \)
- \( d = 1 \), \( e = x \), \( f = 2x^2 \)
- \( g = 1 \), \( h = \frac{1}{x} \), \( i = 2 \)

Plugging these values into the determinant formula gives:

\[ \text{det}(A) = \frac{2}{x}(x \cdot 2 - 2x^2 \cdot \frac{1}{x}) - (-1)(1 \cdot 2 - 2x^2 \cdot 1) + 2(1 \cdot \frac{1}{x} - x \cdot 1) \]

### Step 2: Simplify the Determinant Expression

Calculating each term:

1. First term:
\[ \frac{2}{x}(2x - 2x) = \frac{2}{x}(0) = 0 \]

2. Second term:
\[ -(-1)(2 - 2x^2) = 2 - 2x^2 \]

3. Third term:
\[ 2\left(\frac{1}{x} - x\right) = \frac{2}{x} - 2x \]

Combining these, we have:

\[ \text{det}(A) = 0 + (2 - 2x^2) + \left(\frac{2}{x} - 2x\right) \]

This simplifies to:

\[ \text{det}(A) = 2 - 2x^2 + \frac{2}{x} - 2x \]

### Step 3: Set the Determinant to Zero

To find when the matrix is singular, we set the determinant equal to zero:

\[ 2 - 2x^2 + \frac{2}{x} - 2x = 0 \]

### Step 4: Multiply Through by \( x \) to Eliminate the Fraction

Multiplying the entire equation by \( x \) (assuming \( x \neq 0 \)) gives:

\[ 2x - 2x^3 + 2 - 2x^2 = 0 \]

Rearranging this, we get:

\[ -2x^3 - 2x^2 + 2x + 2 = 0 \]

Dividing through by -2:

\[ x^3 + x^2 - x - 1 = 0 \]

### Step 5: Factor the Polynomial

We can factor this polynomial. We can try \( x = 1 \):

\[ 1^3 + 1^2 - 1 - 1 = 0 \]

So \( x - 1 \) is a factor. We can perform synthetic division or polynomial long division to factor \( x^3 + x^2 - x - 1 \):

\[ x^3 + x^2 - x - 1 = (x - 1)(x^2 + 2x + 1) = (x - 1)(x + 1)^2 \]

### Step 6: Set Each Factor to Zero

Setting each factor to zero gives:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( (x + 1)^2 = 0 \) → \( x = -1 \)

### Final Answer

Thus, the values of \( x \) for which the matrix \( A \) is singular are:

\[ x = 1 \quad \text{and} \quad x = -1 \]