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We are asked to find the value of \( x \) such that \( A^2 = 0 \), where matrix \( A \) is given as:

\[
A = \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix}
\]

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

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

\[
A^2 = A \times A = \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix} \times \begin{bmatrix} 1 & 1 \\ x & x \end{bmatrix}
\]

Performing the matrix multiplication:

\[
A^2 = \begin{bmatrix}
1 \times 1 + 1 \times x & 1 \times 1 + 1 \times x \\
x \times 1 + x \times x & x \times 1 + x \times x
\end{bmatrix}
\]

Simplifying each element:

\[
A^2 = \begin{bmatrix}
1 + x & 1 + x \\
x + x^2 & x + x^2
\end{bmatrix}
\]

### Step 2: Set \( A^2 = 0 \)

We are given that \( A^2 = 0 \), so:

\[
\begin{bmatrix}
1 + x & 1 + x \\
x + x^2 & x + x^2
\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
\]

This gives us two distinct equations from the elements of the matrix:

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

### Step 3: Solve the equations

#### From the first equation:
\[
1 + x = 0 \quad \Rightarrow \quad x = -1
\]

#### From the second equation:
Substitute \( x = -1 \) into \( x + x^2 = 0 \):

\[
-1 + (-1)^2 = 0 \quad \Rightarrow \quad -1 + 1 = 0
\]

This equation holds true.

### Final Answer:
The value of \( x \) is \( \boxed{-1} \).

Would you like more details on any step, or do you have further questions?

### Related Questions:
1. What is the general process for multiplying matrices?
2. How do you check if a matrix is invertible?
3. How can we find the determinant of a 2x2 matrix?
4. What is the geometric interpretation of a matrix product resulting in the zero matrix?
5. How does matrix multiplication relate to systems of equations?

**Tip:** Always double-check your calculations by substituting your solution back into the original equation to verify the result!

If A = [[1, 1], [x, x]], find the value of x so that A^2 = 0.

Find the value of x such that A^2 equals the zero matrix for the given matrix A = [[1, 1], [x, x]]. This problem covers matrix multiplication and solving algebraic equations, suitable for Grades 10-12.

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