1. **Problem 1: Find rank, nullity, and kernel of the linear transformation** Given linear transformation $$T: \mathbb{R}^6 \to \mathbb{R}^4$$ defined by $$T(\mathbf{x}) = A\mathbf{x}$$ where $$A = \begin{pmatrix}-1 & 2 & 0 & 4 & 5 & -3 \\ 3 & -7 & 2 & 0 & 1 & 4 \\ 2 & -5 & 2 & 4 & 6 & 1 \\ 4 & -9 & 2 & -4 & -4 & 7\end{pmatrix}$$ **Step 1:** Find the rank of matrix $A$. Rank is the dimension of the image of $T$, equal to the number of linearly independent rows or columns. Perform row reduction on $A$: $$\begin{pmatrix}-1 & 2 & 0 & 4 & 5 & -3 \\ 3 & -7 & 2 & 0 & 1 & 4 \\ 2 & -5 & 2 & 4 & 6 & 1 \\ 4 & -9 & 2 & -4 & -4 & 7\end{pmatrix}$$ - Multiply row 1 by $-1$: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 3 & -7 & 2 & 0 & 1 & 4 \\ 2 & -5 & 2 & 4 & 6 & 1 \\ 4 & -9 & 2 & -4 & -4 & 7\end{pmatrix}$$ - Row 2 = Row 2 - 3*Row 1: $$3 - 3*1=0, -7 - 3*(-2)= -7 +6 = -1, 2 - 3*0=2, 0 - 3*(-4)=12, 1 - 3*(-5)=16, 4 - 3*3= -5$$ - Row 3 = Row 3 - 2*Row 1: $$2 - 2*1=0, -5 - 2*(-2)= -5 +4 = -1, 2 - 2*0=2, 4 - 2*(-4)=12, 6 - 2*(-5)=16, 1 - 2*3= -5$$ - Row 4 = Row 4 - 4*Row 1: $$4 - 4*1=0, -9 - 4*(-2)= -9 +8 = -1, 2 - 4*0=2, -4 - 4*(-4)=12, -4 - 4*(-5)=16, 7 - 4*3= -5$$ Matrix now: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & -1 & 2 & 12 & 16 & -5\end{pmatrix}$$ - Rows 2, 3, 4 are identical. Replace row 3 and 4 by zeros: Row 3 = Row 3 - Row 2 = 0 Row 4 = Row 4 - Row 2 = 0 Matrix: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$ - Multiply row 2 by $-1$: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 0 & 1 & -2 & -12 & -16 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$ - Row 1 = Row 1 + 2*Row 2: $$1, -2 + 2*1=0, 0 + 2*(-2) = -4, -4 + 2*(-12) = -4 - 24 = -28, -5 + 2*(-16) = -5 - 32 = -37, 3 + 2*5 = 13$$ Final matrix in row echelon form: $$\begin{pmatrix}1 & 0 & -4 & -28 & -37 & 13 \\ 0 & 1 & -2 & -12 & -16 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$$ Rank = number of nonzero rows = 2. **Step 2:** Find nullity using Rank-Nullity Theorem: $$\text{nullity}(T) = \dim(\mathbb{R}^6) - \text{rank}(T) = 6 - 2 = 4$$ **Step 3:** Find kernel $\ker(T)$, the set of all $\mathbf{x}$ such that $A\mathbf{x} = \mathbf{0}$. From the reduced system: $$x_1 - 4x_3 - 28x_4 - 37x_5 + 13x_6 = 0$$ $$x_2 - 2x_3 - 12x_4 - 16x_5 + 5x_6 = 0$$ Express $x_1$ and $x_2$ in terms of free variables $x_3, x_4, x_5, x_6$: $$x_1 = 4x_3 + 28x_4 + 37x_5 - 13x_6$$ $$x_2 = 2x_3 + 12x_4 + 16x_5 - 5x_6$$ Kernel vectors: $$\mathbf{x} = \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \\ x_6\end{pmatrix} = x_3 \begin{pmatrix}4 \\ 2 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix} + x_4 \begin{pmatrix}28 \\ 12 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix} + x_5 \begin{pmatrix}37 \\ 16 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix} + x_6 \begin{pmatrix}-13 \\ -5 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$$ --- 2. **Problem 2: Linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by $T(\mathbf{v}) = A\mathbf{v}$** $$A = \begin{pmatrix}1 & -4 & 2 \\ -2 & 7 & 3 \\ 2 & 6 & 8\end{pmatrix}$$ **Step 1:** Show $T$ is invertible by checking if $\det(A) \neq 0$. Calculate determinant: $$\det(A) = 1 \cdot \begin{vmatrix}7 & 3 \\ 6 & 8\end{vmatrix} - (-4) \cdot \begin{vmatrix}-2 & 3 \\ 2 & 8\end{vmatrix} + 2 \cdot \begin{vmatrix}-2 & 7 \\ 2 & 6\end{vmatrix}$$ Calculate minors: $$= 1(7 \times 8 - 3 \times 6) + 4(-2 \times 8 - 3 \times 2) + 2(-2 \times 6 - 7 \times 2)$$ $$= 1(56 - 18) + 4(-16 - 6) + 2(-12 - 14)$$ $$= 38 + 4(-22) + 2(-26) = 38 - 88 - 52 = -102$$ Since $\det(A) = -102 \neq 0$, $T$ is invertible. **Step 2:** Find $T^{-1} = A^{-1}$ using formula: $$A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ Calculate cofactors and adjugate matrix: Cofactors: $$C_{11} = \begin{vmatrix}7 & 3 \\ 6 & 8\end{vmatrix} = 56 - 18 = 38$$ $$C_{12} = -\begin{vmatrix}-2 & 3 \\ 2 & 8\end{vmatrix} = -(-16 - 6) = 22$$ $$C_{13} = \begin{vmatrix}-2 & 7 \\ 2 & 6\end{vmatrix} = -12 - 14 = -26$$ $$C_{21} = -\begin{vmatrix}-4 & 2 \\ 6 & 8\end{vmatrix} = -(-32 - 12) = 44$$ $$C_{22} = \begin{vmatrix}1 & 2 \\ 2 & 8\end{vmatrix} = 8 - 4 = 4$$ $$C_{23} = -\begin{vmatrix}1 & -4 \\ 2 & 6\end{vmatrix} = -(6 + 8) = -14$$ $$C_{31} = \begin{vmatrix}-4 & 2 \\ 7 & 3\end{vmatrix} = -12 - 14 = -26$$ $$C_{32} = -\begin{vmatrix}1 & 2 \\ -2 & 3\end{vmatrix} = -(3 + 4) = -7$$ $$C_{33} = \begin{vmatrix}1 & -4 \\ -2 & 7\end{vmatrix} = 7 - 8 = -1$$ Adjugate matrix $\text{adj}(A)$ is transpose of cofactor matrix: $$\text{adj}(A) = \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1\end{pmatrix}$$ Therefore, $$A^{-1} = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1\end{pmatrix}$$ **Step 3:** Given $T^{-1}(\mathbf{w}) = \begin{pmatrix}6 \\ 4 \\ 8\end{pmatrix}$, find $\mathbf{w}$. Since $T^{-1}(\mathbf{w}) = \mathbf{v}$, then $\mathbf{w} = T(\mathbf{v}) = A\mathbf{v}$. Calculate: $$\mathbf{w} = A \begin{pmatrix}6 \\ 4 \\ 8\end{pmatrix} = \begin{pmatrix}1 & -4 & 2 \\ -2 & 7 & 3 \\ 2 & 6 & 8\end{pmatrix} \begin{pmatrix}6 \\ 4 \\ 8\end{pmatrix}$$ Calculate each component: $$w_1 = 1 \times 6 + (-4) \times 4 + 2 \times 8 = 6 - 16 + 16 = 6$$ $$w_2 = -2 \times 6 + 7 \times 4 + 3 \times 8 = -12 + 28 + 24 = 40$$ $$w_3 = 2 \times 6 + 6 \times 4 + 8 \times 8 = 12 + 24 + 64 = 100$$ So, $$\mathbf{w} = \begin{pmatrix}6 \\ 40 \\ 100\end{pmatrix}$$ **Step 4:** Given $T(\mathbf{v}) = \begin{pmatrix}8 \\ 4 \\ 9\end{pmatrix}$, find $\mathbf{v}$. Since $T(\mathbf{v}) = A\mathbf{v} = \mathbf{b}$, then $$\mathbf{v} = A^{-1} \mathbf{b}$$ Calculate: $$\mathbf{v} = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1\end{pmatrix} \begin{pmatrix}8 \\ 4 \\ 9\end{pmatrix}$$ Multiply: $$\begin{pmatrix}38 \times 8 + 44 \times 4 + (-26) \times 9 \\ 22 \times 8 + 4 \times 4 + (-7) \times 9 \\ -26 \times 8 + (-14) \times 4 + (-1) \times 9\end{pmatrix} = \begin{pmatrix}304 + 176 - 234 \\ 176 + 16 - 63 \\ -208 - 56 - 9\end{pmatrix} = \begin{pmatrix}246 \\ 129 \\ -273\end{pmatrix}$$ Divide by $-102$: $$\mathbf{v} = \begin{pmatrix} \frac{246}{-102} \\ \frac{129}{-102} \\ \frac{-273}{-102} \end{pmatrix} = \begin{pmatrix} -\frac{123}{51} \\ -\frac{43}{34} \\ \frac{91}{34} \end{pmatrix}$$ Simplify fractions: $$\mathbf{v} = \begin{pmatrix} -\frac{41}{17} \\ -\frac{43}{34} \\ \frac{91}{34} \end{pmatrix}$$ --- **Final answers:** 1) a) Rank = 2 b) Nullity = 4 c) Kernel basis vectors: $$\left\{ \begin{pmatrix}4 \\ 2 \\ 1 \\ 0 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}28 \\ 12 \\ 0 \\ 1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}37 \\ 16 \\ 0 \\ 0 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix}-13 \\ -5 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix} \right\}$$ 2) a) $T$ is invertible since $\det(A) = -102 \neq 0$ b) $$T^{-1} = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1\end{pmatrix}$$ c) $$\mathbf{w} = \begin{pmatrix}6 \\ 40 \\ 100\end{pmatrix}$$ d) $$\mathbf{v} = \begin{pmatrix} -\frac{41}{17} \\ -\frac{43}{34} \\ \frac{91}{34} \end{pmatrix}$$