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 matrix $$A = \begin{pmatrix}-1 & 2 & 0 & 4 & 5 & -3 \\ 3 & -7 & 2 & 0 & 1 & 4 \\ 2 & -5 & 2 & 4 & 6 & 1 \\ 4 & -9 & 2 & -4 & -4 & 7 \end{pmatrix}$$ We want to find: - a) Rank of $T$ - b) Nullity of $T$ - c) Kernel of $T$ --- **Step 1: Find the rank of $T$ (rank of matrix $A$)** Rank is the dimension of the image of $T$, equal to the number of linearly independent rows or columns. We perform row reduction on $A$ to find its rank. $$A = \begin{pmatrix}-1 & 2 & 0 & 4 & 5 & -3 \\ 3 & -7 & 2 & 0 & 1 & 4 \\ 2 & -5 & 2 & 4 & 6 & 1 \\ 4 & -9 & 2 & -4 & -4 & 7 \end{pmatrix}$$ - Multiply row 1 by $-1$ for easier calculations: $$R_1 \to -R_1 = (1, -2, 0, -4, -5, 3)$$ - Use $R_1$ to eliminate first column entries in $R_2, R_3, R_4$: $$R_2 \to R_2 - 3R_1 = (3 - 3*1, -7 - 3*(-2), 2 - 3*0, 0 - 3*(-4), 1 - 3*(-5), 4 - 3*3) = (0, -7 + 6, 2, 0 + 12, 1 + 15, 4 - 9) = (0, -1, 2, 12, 16, -5)$$ $$R_3 \to R_3 - 2R_1 = (2 - 2*1, -5 - 2*(-2), 2 - 2*0, 4 - 2*(-4), 6 - 2*(-5), 1 - 2*3) = (0, -5 + 4, 2, 4 + 8, 6 + 10, 1 - 6) = (0, -1, 2, 12, 16, -5)$$ $$R_4 \to R_4 - 4R_1 = (4 - 4*1, -9 - 4*(-2), 2 - 4*0, -4 - 4*(-4), -4 - 4*(-5), 7 - 4*3) = (0, -9 + 8, 2, -4 + 16, -4 + 20, 7 - 12) = (0, -1, 2, 12, 16, -5)$$ - Now matrix is: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & -1 & 2 & 12 & 16 & -5 \end{pmatrix}$$ - Rows 2, 3, 4 are identical. Subtract $R_2$ from $R_3$ and $R_4$: $$R_3 \to R_3 - R_2 = 0$$ $$R_4 \to R_4 - R_2 = 0$$ - Matrix reduces to: $$\begin{pmatrix}1 & -2 & 0 & -4 & -5 & 3 \\ 0 & -1 & 2 & 12 & 16 & -5 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}$$ - Rank is number of nonzero rows = 2. **Answer a):** $\boxed{\text{rank}(T) = 2}$ --- **Step 2: Find nullity of $T$** By the Rank-Nullity Theorem: $$\text{nullity}(T) = \dim(\mathbb{R}^6) - \text{rank}(T) = 6 - 2 = 4$$ **Answer b):** $\boxed{\text{nullity}(T) = 4}$ --- **Step 3: Find kernel of $T$ (null space of $A$)** Solve $A\mathbf{x} = \mathbf{0}$ for $\mathbf{x} = (x_1, x_2, x_3, x_4, x_5, x_6)^T$. From row-reduced form: 1) $x_1 - 2x_2 - 4x_4 - 5x_5 + 3x_6 = 0$ 2) $-x_2 + 2x_3 + 12x_4 + 16x_5 - 5x_6 = 0$ Express leading variables $x_1, x_2$ in terms of free variables $x_3, x_4, x_5, x_6$: From (2): $$x_2 = 2x_3 + 12x_4 + 16x_5 - 5x_6$$ Substitute into (1): $$x_1 - 2(2x_3 + 12x_4 + 16x_5 - 5x_6) - 4x_4 - 5x_5 + 3x_6 = 0$$ Simplify: $$x_1 - 4x_3 - 24x_4 - 32x_5 + 10x_6 - 4x_4 - 5x_5 + 3x_6 = 0$$ $$x_1 - 4x_3 - 28x_4 - 37x_5 + 13x_6 = 0$$ So: $$x_1 = 4x_3 + 28x_4 + 37x_5 - 13x_6$$ **Kernel basis vectors:** Set each free variable to 1 one at a time, others 0: - For $x_3=1$, others 0: $$x_1=4, x_2=2, x_3=1, x_4=0, x_5=0, x_6=0$$ - For $x_4=1$, others 0: $$x_1=28, x_2=12, x_3=0, x_4=1, x_5=0, x_6=0$$ - For $x_5=1$, others 0: $$x_1=37, x_2=16, x_3=0, x_4=0, x_5=1, x_6=0$$ - For $x_6=1$, others 0: $$x_1=-13, x_2=-5, x_3=0, x_4=0, x_5=0, x_6=1$$ Thus, $$\ker(T) = \text{span}\left\{ \begin{pmatrix}4 \\ 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix}28 \\ 12 \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix}37 \\ 16 \\ 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix}-13 \\ -5 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}$$ **Answer c):** Kernel is the 4-dimensional subspace spanned by above vectors. --- 2. **Problem 2: Linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by matrix** $$A = \begin{pmatrix}1 & -4 & 2 \\ -2 & 7 & 3 \\ 2 & 6 & 8 \end{pmatrix}$$ We want to find: - a) Show $T$ is invertible - b) Find $T^{-1}$ formula - c) If $T^{-1}(w) = (6,4,8)$, find $w$ - d) If $T(v) = (8,4,9)$, find $v$ --- **Step 1: Check invertibility by determinant** Calculate $\det(A)$: $$\det(A) = 1 \cdot \begin{vmatrix}7 & 3 \\ 6 & 8 \end{vmatrix} - (-4) \cdot \begin{vmatrix}-2 & 3 \\ 2 & 8 \end{vmatrix} + 2 \cdot \begin{vmatrix}-2 & 7 \\ 2 & 6 \end{vmatrix}$$ 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 $\det(A) \neq 0$, $T$ is invertible. **Answer a):** $T$ is invertible because $\det(A) = -102 \neq 0$. --- **Step 2: Find $T^{-1}$** Formula: $$T^{-1} = A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$$ Calculate adjugate matrix $\text{adj}(A)$ by cofactors: Cofactors matrix $C$: $$C = \begin{pmatrix}C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix}$$ Calculate each cofactor: - $C_{11} = \det \begin{pmatrix}7 & 3 \\ 6 & 8 \end{pmatrix} = 56 - 18 = 38$ - $C_{12} = -\det \begin{pmatrix}-2 & 3 \\ 2 & 8 \end{pmatrix} = -(-16 - 6) = 22$ - $C_{13} = \det \begin{pmatrix}-2 & 7 \\ 2 & 6 \end{pmatrix} = -12 - 14 = -26$ - $C_{21} = -\det \begin{pmatrix}-4 & 2 \\ 6 & 8 \end{pmatrix} = -(-32 - 12) = 44$ - $C_{22} = \det \begin{pmatrix}1 & 2 \\ 2 & 8 \end{pmatrix} = 8 - 4 = 4$ - $C_{23} = -\det \begin{pmatrix}1 & -4 \\ 2 & 6 \end{pmatrix} = -(6 + 8) = -14$ - $C_{31} = \det \begin{pmatrix}-4 & 2 \\ 7 & 3 \end{pmatrix} = -12 - 14 = -26$ - $C_{32} = -\det \begin{pmatrix}1 & 2 \\ -2 & 3 \end{pmatrix} = -(3 + 4) = -7$ - $C_{33} = \det \begin{pmatrix}1 & -4 \\ -2 & 7 \end{pmatrix} = 7 - 8 = -1$ So, $$C = \begin{pmatrix}38 & 22 & -26 \\ 44 & 4 & -14 \\ -26 & -7 & -1 \end{pmatrix}$$ Adjugate is transpose of $C$: $$\text{adj}(A) = C^T = \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1 \end{pmatrix}$$ Therefore, $$A^{-1} = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1 \end{pmatrix}$$ **Answer b):** $$T^{-1}(\mathbf{y}) = A^{-1} \mathbf{y} = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1 \end{pmatrix} \mathbf{y}$$ --- **Step 3: Find $w$ if $T^{-1}(w) = (6,4,8)$** Given: $$T^{-1}(w) = (6,4,8)$$ Apply $T$ to both sides: $$w = T(6,4,8) = A \begin{pmatrix}6 \\ 4 \\ 8 \end{pmatrix}$$ Calculate: $$w = \begin{pmatrix}1 & -4 & 2 \\ -2 & 7 & 3 \\ 2 & 6 & 8 \end{pmatrix} \begin{pmatrix}6 \\ 4 \\ 8 \end{pmatrix} = \begin{pmatrix}1*6 + (-4)*4 + 2*8 \\ -2*6 + 7*4 + 3*8 \\ 2*6 + 6*4 + 8*8 \end{pmatrix} = \begin{pmatrix}6 - 16 + 16 \\ -12 + 28 + 24 \\ 12 + 24 + 64 \end{pmatrix} = \begin{pmatrix}6 \\ 40 \\ 100 \end{pmatrix}$$ **Answer c):** $\boxed{w = (6, 40, 100)}$ --- **Step 4: Find $v$ if $T(v) = (8,4,9)$** We want $v = T^{-1}(8,4,9) = A^{-1} \begin{pmatrix}8 \\ 4 \\ 9 \end{pmatrix}$ Calculate: $$v = \frac{1}{-102} \begin{pmatrix}38 & 44 & -26 \\ 22 & 4 & -7 \\ -26 & -14 & -1 \end{pmatrix} \begin{pmatrix}8 \\ 4 \\ 9 \end{pmatrix} = \frac{1}{-102} \begin{pmatrix}38*8 + 44*4 - 26*9 \\ 22*8 + 4*4 - 7*9 \\ -26*8 - 14*4 - 1*9 \end{pmatrix}$$ Calculate each component: - First: $$38*8 = 304, \quad 44*4 = 176, \quad -26*9 = -234$$ $$304 + 176 - 234 = 246$$ - Second: $$22*8 = 176, \quad 4*4 = 16, \quad -7*9 = -63$$ $$176 + 16 - 63 = 129$$ - Third: $$-26*8 = -208, \quad -14*4 = -56, \quad -1*9 = -9$$ $$-208 - 56 - 9 = -273$$ So, $$v = \frac{1}{-102} \begin{pmatrix}246 \\ 129 \\ -273 \end{pmatrix} = \begin{pmatrix}-\frac{246}{102} \\ -\frac{129}{102} \\ \frac{273}{102} \end{pmatrix}$$ Simplify fractions: - $\frac{246}{102} = \frac{123}{51} = \frac{41}{17}$ - $\frac{129}{102} = \frac{43}{34}$ - $\frac{273}{102} = \frac{91}{34}$ Final: $$v = \begin{pmatrix}-\frac{41}{17} \\ -\frac{43}{34} \\ \frac{91}{34} \end{pmatrix}$$ **Answer d):** $v = \left(-\frac{41}{17}, -\frac{43}{34}, \frac{91}{34}\right)$ --- **Summary:** - Problem 1: - a) rank = 2 - b) nullity = 4 - c) kernel basis as above - Problem 2: - a) invertible since $\det(A) = -102 \neq 0$ - b) inverse matrix formula given - c) $w = (6, 40, 100)$ - d) $v = \left(-\frac{41}{17}, -\frac{43}{34}, \frac{91}{34}\right)$