You are given the matrix $A$:

$A = \begin{pmatrix} 4 & -12 \\ -2 & 6 \end{pmatrix}$

You are tasked with constructing a $2 \times 2$ matrix $B$ such that:

$AB = 0$

This implies that the product of $A$ and $B$ must result in the zero matrix. Let's denote $B$ as:

$B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}$

Now, we compute $AB$:

$AB = \begin{pmatrix} 4 & -12 \\ -2 & 6 \end{pmatrix} \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} = \begin{pmatrix} 4b_{11} - 12b_{21} & 4b_{12} - 12b_{22} \\ -2b_{11} + 6b_{21} & -2b_{12} + 6b_{22} \end{pmatrix}$

For this product to be the zero matrix, we need the following system of equations:

4b_{11} - 12b_{21} &= 0 \\ 4b_{12} - 12b_{22} &= 0 \\ -2b_{11} + 6b_{21} &= 0 \\ -2b_{12} + 6b_{22} &= 0 \end{aligned}$$ Simplifying the equations: 1. $$4b_{11} - 12b_{21} = 0 \quad \Rightarrow \quad b_{11} = 3b_{21}$$ 2. $$4b_{12} - 12b_{22} = 0 \quad \Rightarrow \quad b_{12} = 3b_{22}$$ 3. $$-2b_{11} + 6b_{21} = 0 \quad \Rightarrow \quad b_{11} = 3b_{21}$$ (same as equation 1) 4. $$-2b_{12} + 6b_{22} = 0 \quad \Rightarrow \quad b_{12} = 3b_{22}$$ (same as equation 2) Thus, the two columns of $$B$$ must satisfy $$b_{11} = 3b_{21}$$ and $$b_{12} = 3b_{22}$$. Therefore, we can choose two different non-zero columns for $$B$$. Here are two examples of $$B$$: 1. $$B = \begin{pmatrix} 3 & 0 \\ 1 & 0 \end{pmatrix}$$ 2. $$B = \begin{pmatrix} 0 & 3 \\ 0 & 1 \end{pmatrix}$$ Both choices satisfy the condition that $$AB = 0$$, and they use different nonzero columns for $$B$$. Would you like further details or have any questions? Here are five related questions: 1. How do we generally compute the product of two matrices? 2. What is the significance of the zero matrix in matrix equations? 3. Can a matrix have multiple solutions for $$B$$ if $$AB = 0$$? 4. What does it mean for a matrix to be invertible? 5. How does the determinant of a matrix relate to its invertibility? **Tip**: In linear algebra, a homogeneous system of equations often leads to non-trivial solutions if the matrix is singular (i.e., not invertible).