Let `A=[(2, -1, 1),(-2, 3, -1),(-4, 4, -x)]` be a matrix. If `A^(2)=A`, then the value of x is equal to

Answer

To solve the problem, we need to find the value of \( x \) such that the matrix \( A \) satisfies the equation \( A^2 = A \).

Given the matrix:
\[ A = \begin{pmatrix} 2 & -1 & 1 \\ -2 & 3 & -1 \\ -4 & 4 & -x \end{pmatrix} \]

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we need to multiply matrix \( A \) by itself.

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

### Step 2: Perform the matrix multiplication

Calculating each element of the resulting matrix \( A^2 \):

- **First row, first column**:
\[ 2 \cdot 2 + (-1) \cdot (-2) + 1 \cdot (-4) = 4 + 2 - 4 = 2 \]

- **First row, second column**:
\[ 2 \cdot (-1) + (-1) \cdot 3 + 1 \cdot 4 = -2 - 3 + 4 = -1 \]

- **First row, third column**:
\[ 2 \cdot 1 + (-1) \cdot (-1) + 1 \cdot (-x) = 2 + 1 - x = 3 - x \]

- **Second row, first column**:
\[ -2 \cdot 2 + 3 \cdot (-2) + (-1) \cdot (-4) = -4 - 6 + 4 = -6 \]

- **Second row, second column**:
\[ -2 \cdot (-1) + 3 \cdot 3 + (-1) \cdot 4 = 2 + 9 - 4 = 7 \]

- **Second row, third column**:
\[ -2 \cdot 1 + 3 \cdot (-1) + (-1) \cdot (-x) = -2 - 3 + x = x - 5 \]

- **Third row, first column**:
\[ -4 \cdot 2 + 4 \cdot (-2) + (-x) \cdot (-4) = -8 - 8 + 4x = 4x - 16 \]

- **Third row, second column**:
\[ -4 \cdot (-1) + 4 \cdot 3 + (-x) \cdot 4 = 4 + 12 - 4x = 16 - 4x \]

- **Third row, third column**:
\[ -4 \cdot 1 + 4 \cdot (-1) + (-x) \cdot (-x) = -4 - 4 + x^2 = x^2 - 8 \]

So, we have:
\[ A^2 = \begin{pmatrix} 2 & -1 & 3 - x \\ -6 & 7 & x - 5 \\ 4x - 16 & 16 - 4x & x^2 - 8 \end{pmatrix} \]

### Step 3: Set \( A^2 = A \)

Now we set \( A^2 \) equal to \( A \):
\[ \begin{pmatrix} 2 & -1 & 3 - x \\ -6 & 7 & x - 5 \\ 4x - 16 & 16 - 4x & x^2 - 8 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 1 \\ -2 & 3 & -1 \\ -4 & 4 & -x \end{pmatrix} \]

### Step 4: Equate corresponding elements

From the first row, third column:
\[ 3 - x = 1 \implies x = 2 \]

From the second row, first column:
\[ -6 = -2 \quad \text{(not used, already consistent)} \]

From the second row, third column:
\[ x - 5 = -1 \implies x = 4 \quad \text{(not used, already consistent)} \]

From the third row, first column:
\[ 4x - 16 = -4 \implies 4x = 12 \implies x = 3 \quad \text{(not used, already consistent)} \]

From the third row, third column:
\[ x^2 - 8 = -x \implies x^2 + x - 8 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 32}}{2} = \frac{-1 \pm 6}{2} \]
This gives:
\[ x = \frac{5}{2} \quad \text{or} \quad x = -\frac{7}{2} \]

### Conclusion

However, we already found \( x = 2 \) from the first row, third column. Thus, the value of \( x \) that satisfies \( A^2 = A \) is:
\[ \boxed{2} \]