(8) (15 points) Let \( L: R^{4} \rightarrow R^{3} \) be the linear transformation defined by \( L(\vec{x})=\left[\begin{array}{cccc}1 & 2 & -3 & 12 \\ 2 & 4 & 2 & 0 \\ 3 & 6 & 1 & 6\end{array}\right] \cdot \vec{x} \). (a) Find a basis for the kernel of \( L \). (b) Find a basis for the range (image) of \( L \).

We are given a linear transformation L from ℝ⁴ to ℝ³ defined by   L(x) = A·x  with       A = | 1  2  –3  12 |       | 2  4   2   0 |       | 3  6   1   6 | where x = (x₁, x₂, x₃, x₄)ᵀ. We wish to find:  (a) A basis for the kernel of L (i.e. all vectors x ∈ ℝ⁴ with L(x) = 0).  (b) A basis for the range (or image) of L (that is, the span of the columns of A). ──────────────────────────── Step (a): Find a Basis for the Kernel of L The kernel is defined by the homogeneous system   1) x₁ + 2x₂ – 3x₃ + 12x₄ = 0   2) 2x₁ + 4x₂ + 2x₃ = 0   3) 3x₁ + 6x₂ + x₃ + 6x₄ = 0 Begin with equation (2). Dividing by 2 gives   x₁ + 2x₂ + x₃ = 0    → x₁ = –2x₂ – x₃. Now substitute this expression for x₁ into equation (1):   (–2x₂ – x₃) + 2x₂ – 3x₃ + 12x₄ = 0     – x₃ – 3x₃ + 12x₄ = 0     –4x₃ + 12x₄ = 0    → x₃ = 3x₄. Now substitute x₃ = 3x₄ back into the expression for x₁:   x₁ = –2x₂ – (3x₄). Finally, check equation (3). Substitute x₁ and x₃ into (3):   3(–2x₂ – 3x₄) + 6x₂ + (3x₄) + 6x₄ =     –6x₂ – 9x₄ + 6x₂ + 3x₄ + 6x₄ = 0. It is automatically satisfied, so there is no further restriction. Thus, we have two free parameters: Let x₂ = s and x₄ = t, where s, t ∈ ℝ. Then   x₃ = 3t, and x₁ = –2s – 3t. So the general solution is   x = (x₁, x₂, x₃, x₄) = (–2s – 3t, s, 3t, t)          = s·(–2, 1, 0, 0) + t·(–3, 0, 3, 1). A basis for the kernel is therefore   { (–2, 1, 0, 0),  (–3, 0, 3, 1) }. ──────────────────────────── Step (b): Find a Basis for the Range (Image) of L The range of L is the column space of A. The columns of A are   c₁ = (1, 2, 3),    c₂ = (2, 4, 6),    c₃ = (–3, 2, 1),    c₄ = (12, 0, 6). Notice that   c₂ = 2·c₁    (so c₂ is not independent). To find the remaining independent columns, we can perform row reduction or check for linear dependence between c₁, c₃, and c₄. One way is to note that the row reduction of A reveals that A has rank 2. (In fact, if we subtract 2 times row 1 from row 2 and 3 times row 1 from row 3, we find that there are only two pivot rows.) This indicates that we need only 2 independent columns. We can choose as our basis the columns corresponding to the pivot columns. For example, one may take c₁ and c₃. (One can also verify that c₄ is a linear combination of c₁ and c₃.) Thus, a basis for the range (image) of L is   { (1, 2, 3), (–3, 2, 1) }. ──────────────────────────── Final Answers (a) A basis for the kernel of L is   { (–2, 1, 0, 0), (–3, 0, 3, 1) }. (b) A basis for the range (image) of L is   { (1, 2, 3), (–3, 2, 1) }. These answers complete the problem.