To find a basis for the kernel of $A$, we solve the equation:

$A \mathbf{x} = 0$

where $A = \begin{bmatrix} -1 & -2 & 6 \\ 1 & 2 & -6 \\ 0 & 0 & 0 \end{bmatrix}$ and $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$.

Step 1: Write $A \mathbf{x} = 0$ as a system of equations

Expanding $A \mathbf{x} = 0$, we get:

- x_1 - 2x_2 + 6x_3 &= 0, \\ x_1 + 2x_2 - 6x_3 &= 0, \\ 0 &= 0. \end{aligned}$$ The third equation is redundant, so we only need to solve the first two equations. ### Step 2: Eliminate redundancy in the system Adding the first and second equations: $$(-x_1 - 2x_2 + 6x_3) + (x_1 + 2x_2 - 6x_3) = 0 \implies 0 = 0.$$ Thus, the system reduces to a single independent equation: $$- x_1 - 2x_2 + 6x_3 = 0.$$ Rewriting this: $$x_1 = -2x_2 + 6x_3.$$ ### Step 3: Parameterize the solution Let $$x_2 = t$$ and $$x_3 = s$$, where $$t, s \in \mathbb{R}$$. Then: $$x_1 = -2t + 6s.$$ Thus, the general solution is: $$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2t + 6s \\ t \\ s \end{bmatrix} = t \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 6 \\ 0 \\ 1 \end{bmatrix}.$$ ### Step 4: Identify the basis of $$\ker(A)$$ The kernel is spanned by the vectors: $$\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \quad \begin{bmatrix} 6 \\ 0 \\ 1 \end{bmatrix}.$$ Thus, a basis for $$\ker(A)$$ is: $$\boxed{\left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 6 \\ 0 \\ 1 \end{bmatrix} \right\}}.$$ Would you like me to verify this or provide more details? ### Related Questions: 1. What is the rank of $$A$$ based on this computation? 2. How can we confirm that $$B$$ is invertible? 3. What is the relationship between the row space and the null space of $$A$$? 4. If $$A$$ were squared, how would its kernel change? 5. What is the dimension of the null space of $$A$$? **Tip:** Always check that the vectors in your kernel basis are linearly independent to confirm they span the correct null space.