If `A=[(2,-2,-4),(-1,3,4),(1,-2,x)]` is an idempotent matrix, then x=………….

Answer

To find the value of \( x \) in the idempotent matrix \( A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{pmatrix} \), we need to use the property of idempotent matrices, which states that \( A^2 = A \).

### Step 1: Calculate \( A^2 \)
First, we calculate \( A^2 \) by multiplying matrix \( A \) by itself.

\[ A^2 = A \cdot A = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{pmatrix} \cdot \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{pmatrix} \]

### Step 2: Perform the Matrix Multiplication
Now, we will multiply the two matrices. The element in the \( i^{th} \) row and \( j^{th} \) column of the resulting matrix is obtained by taking the dot product of the \( i^{th} \) row of the first matrix and the \( j^{th} \) column of the second matrix.

1. **First Row:**
- First element: \( 2 \cdot 2 + (-2) \cdot (-1) + (-4) \cdot 1 = 4 + 2 - 4 = 2 \)
- Second element: \( 2 \cdot (-2) + (-2) \cdot 3 + (-4) \cdot (-2) = -4 - 6 + 8 = -2 \)
- Third element: \( 2 \cdot (-4) + (-2) \cdot 4 + (-4) \cdot x = -8 - 8 - 4x = -16 - 4x \)

2. **Second Row:**
- First element: \( -1 \cdot 2 + 3 \cdot (-1) + 4 \cdot 1 = -2 - 3 + 4 = -1 \)
- Second element: \( -1 \cdot (-2) + 3 \cdot 3 + 4 \cdot (-2) = 2 + 9 - 8 = 3 \)
- Third element: \( -1 \cdot (-4) + 3 \cdot 4 + 4 \cdot x = 4 + 12 + 4x = 16 + 4x \)

3. **Third Row:**
- First element: \( 1 \cdot 2 + (-2) \cdot (-1) + x \cdot 1 = 2 + 2 + x = 4 + x \)
- Second element: \( 1 \cdot (-2) + (-2) \cdot 3 + x \cdot (-2) = -2 - 6 - 2x = -8 - 2x \)
- Third element: \( 1 \cdot (-4) + (-2) \cdot 4 + x \cdot x = -4 - 8 + x^2 = x^2 - 12 \)

Combining these results, we have:

\[ A^2 = \begin{pmatrix} 2 & -2 & -16 - 4x \\ -1 & 3 & 16 + 4x \\ 4 + x & -8 - 2x & x^2 - 12 \end{pmatrix} \]

### Step 3: Set \( A^2 = A \)
Now, we set \( A^2 \) equal to \( A \):

\[ \begin{pmatrix} 2 & -2 & -16 - 4x \\ -1 & 3 & 16 + 4x \\ 4 + x & -8 - 2x & x^2 - 12 \end{pmatrix} = \begin{pmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & x \end{pmatrix} \]

### Step 4: Solve the Equations
From the equality of matrices, we can derive the following equations:

1. From the first row, third column:
\[ -16 - 4x = -4 \implies -4x = 12 \implies x = -3 \]

2. From the second row, third column:
\[ 16 + 4x = 4 \implies 4x = -12 \implies x = -3 \]

3. From the third row, first column:
\[ 4 + x = 1 \implies x = -3 \]

4. From the third row, second column:
\[ -8 - 2x = -2 \implies -2x = 6 \implies x = -3 \]

5. From the third row, third column:
\[ x^2 - 12 = x \implies x^2 - x - 12 = 0 \]
Factoring gives:
\[ (x - 4)(x + 3) = 0 \implies x = 4 \text{ or } x = -3 \]

### Conclusion
Since all equations consistently lead us to \( x = -3 \), we conclude:

\[ \boxed{-3} \]