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To solve the problem, we need to find a non-zero \(2 \times 2\) matrix \( B \) such that \( AB = C \), where:
\[ A = \begin{pmatrix} -8 & -1 \\ 24 & 3 \end{pmatrix} \]
\[ C = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

Let \( B = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).

We need to perform the matrix multiplication \( AB \) and set it equal to \( C \).

1. Multiply \( A \) by \( B \):
\[ AB = \begin{pmatrix} -8 & -1 \\ 24 & 3 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} -8a - c & -8b - d \\ 24a + 3c & 24b + 3d \end{pmatrix} \]

2. Set \( AB \) equal to \( C \):
\[ \begin{pmatrix} -8a - c & -8b - d \\ 24a + 3c & 24b + 3d \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

This gives us a system of linear equations:
\[
\begin{cases}
-8a - c = 0 \\
-8b - d = 0 \\
24a + 3c = 0 \\
24b + 3d = 0
\end{cases}
\]

3. Solve the system of equations:

From the first equation:
\[ -8a - c = 0 \implies c = -8a \]

From the second equation:
\[ -8b - d = 0 \implies d = -8b \]

Substitute \( c \) in the third equation:
\[ 24a + 3(-8a) = 0 \implies 24a - 24a = 0 \implies 0 = 0 \]
This is always true and does not provide new information.

Substitute \( d \) in the fourth equation:
\[ 24b + 3(-8b) = 0 \implies 24b - 24b = 0 \implies 0 = 0 \]
This is also always true and does not provide new information.

Therefore, we have:
\[ c = -8a \]
\[ d = -8b \]

So, \( B \) can be any non-zero matrix of the form:
\[ B = \begin{pmatrix} a & b \\ -8a & -8b \end{pmatrix} \]

For example, we can choose:
\[ a = 1 \]
\[ b = 1 \]

Then:
\[ c = -8 \times 1 = -8 \]
\[ d = -8 \times 1 = -8 \]

Thus, one possible solution for \( B \) is:
\[ B = \begin{pmatrix} 1 & 1 \\ -8 & -8 \end{pmatrix} \]

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. How do you perform matrix multiplication?
2. What are the conditions for matrix multiplication to be defined?
3. How can you verify if a given matrix \( B \) satisfies \( AB = C \)?
4. What are some properties of the zero matrix \( C \) in matrix equations?
5. How would the solution change if \( C \) were a non-zero matrix?

**Tip:** When solving systems of linear equations derived from matrix equations, always check if the equations provide independent constraints to avoid redundant information.

Find a non-zero 2x2 matrix B such that AB = C, where A = [[-8, -1], [24, 3]] and C = [[0, 0], [0, 0]].

Learn how to solve for a non-zero 2x2 matrix B such that AB = C, where A = [[-8, -1], [24, 3]] and C is a zero matrix. This problem involves matrix multiplication and solving a system of linear equations, suitable for undergraduate students.

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