Let's solve this step by step for **matrix A** first, and explain after each step. Matrix A: \[ A = \begin{bmatrix} 1 & 2 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 2 \end{bmatrix} \] --- ## Step 1: **Linear Dependence/Independence & Column Space Basis** ### (a) Are the columns of the matrix linearly dependent or independent? Notice that matrix A is upper triangular with nonzero diagonal entries (1, 1, 2). In such cases, the columns are **linearly independent**. **Explanation:** A square matrix with nonzero entries on the diagonal and zeros below is invertible, and thus its columns are linearly independent. ### (b) Find the basis for the column space (ColA). Since all columns are linearly independent, the basis for the column space is just the set of all columns: \[ \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 6 \\ 4 \\ 2 \end{bmatrix} \right\} \] **Explanation:** The column space basis is simply the set of independent columns of the matrix. --- ## Step 2: **Null Space Basis, Rank, Nullity, and Rank-Nullity Theorem** ### (c) Find the basis for the null space (NullA). Solve \( A\vec{x} = 0 \): \[ \begin{bmatrix} 1 & 2 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \] Start from the bottom: - Row 3: \(2x_3 = 0 \implies x_3 = 0\) - Row 2: \(x_2 + 4x_3 = 0 \implies x_2 = 0\) - Row 1: \(x_1 + 2x_2 + 6x_3 = 0 \implies x_1 = 0\) So, the only solution is the zero vector. **Null space is just \(\{0\}\), so its basis is the empty set.** **Explanation:** No free variables in the homogeneous system, so only the trivial solution exists; the null space only contains the zero vector. --- ### (d) Rank, Nullity, and Rank-Nullity Theorem - **Rank:** Number of pivot columns = 3 - **Nullity:** Number of free variables = 0 - **Rank-Nullity Theorem:** For \(3 \times 3\) matrix, \( \text{Rank} + \text{Nullity} = 3 \) - Here, \(3 + 0 = 3\), so the theorem is confirmed. **Explanation:** Rank-Nullity Theorem states that for any matrix, the dimension of the column space plus the dimension of the null space equals the number of columns. --- ## **Final Answers for Matrix A:** 1. Columns are linearly independent. 2. Column space basis: \( \{ [1\ 0\ 0]^T, [2\ 1\ 0]^T, [6\ 4\ 2]^T \} \) 3. Null space basis: \( \varnothing \) (only the zero vector). 4. Rank = 3, Nullity = 0. Rank-Nullity Theorem is confirmed. --- Would you like to proceed similarly for matrices B and C?

Let's analyze each matrix step by step, following the same approach as with matrix A. --- ## Matrix A: \[ A = \begin{bmatrix} 1 & 7 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \] --- ### **(a) Are the columns linearly dependent or independent?** **Explanation:** To determine linear dependence/independence, check the rank of the matrix or whether columns can be expressed as linear combinations of each other. Since the matrix is 4x3, and the leading entries in the rows are non-zero (the matrix appears to be full rank), the columns are likely linearly independent. **Step:** Perform row operations or check the rank. --- ### **(b) Find the basis for the column space of A** **Explanation:** The basis for the column space consists of the columns corresponding to pivot columns in the row-reduced form. Since the matrix appears to be in upper triangular form with pivots in columns 1, 2, and 3, all columns are pivot columns. Thus, the basis for the column space is the set of all columns: \[ \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 7 \\ 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 6 \\ 4 \\ 1 \\ 3 \end{bmatrix} \right\} \] --- ### **(c) Find the basis for the null space of A** **Explanation:** Solve \(A \vec{x} = 0\). The number of free variables equals \( \text{nullity} = \text{columns} - \text{rank} \). Given the structure, the rank is 3 (full rank), since there are 3 pivot positions, and there are 4 columns. - Number of variables: 3 - Rank: 3 - Nullity: \(4 - 3 = 1\) Find the null space basis: Set the augmented matrix in row echelon form and solve: \[ \begin{bmatrix} 1 & 7 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \] Subtract the first row from the last: Row 4: \( R_4 - R_1 \rightarrow [0, -5, -3] \) Now, the system corresponds to: \[ \begin{cases} x_1 + 7x_2 + 6x_3 = 0 \\ x_2 + 4x_3 = 0 \\ x_3 = 0 \end{cases} \] From the third row: \(x_3 = 0\) From the second row: \(x_2 + 4 \times 0 = 0 \Rightarrow x_2 = 0\) From the first row: \(x_1 + 7 \times 0 + 6 \times 0 = 0 \Rightarrow x_1 = 0\) The null space contains only the zero vector; thus, the null space basis is empty. **Note:** Since the rank is 3, the null space is trivial. --- ### **(d) Rank, nullity, and Rank-Nullity Theorem** - **Rank:** 3 - **Nullity:** 1 (since 4 columns, rank 3) - **Verification:** \( \text{rank} + \text{nullity} = 3 + 1 = 4 \), which equals the number of columns. --- ## **Final Summary for Matrix A:** - (a) Columns are linearly independent. - (b) Column space basis: \(\left\{ \text{columns 1, 2, 3} \right\}\) - (c) Null space basis: empty set (trivial null space). - (d) Rank = 3, Nullity = 1, Rank-Nullity Theorem confirmed. --- Now, proceed with **Matrix B**: \[ B = \begin{bmatrix} 1 & 4 & 2 \\ -1 & 2 & -2 \\ 3 & 3 & 6 \end{bmatrix} \] --- ### **(a) Are the columns linearly dependent or independent?** **Explanation:** Calculate the rank or check for linear dependence. Perform row operations: - Use row 1 as pivot: \[ R_2 + R_1: \quad (-1 + 1, 2 + 4, -2 + 2) = (0, 6, 0) \] - \( R_3 - 3 \times R_1: \quad (3 - 3, 3 - 12, 6 - 6) = (0, -9, 0) \) Now, the matrix becomes: \[ \begin{bmatrix} 1 & 4 & 2 \\ 0 & 6 & 0 \\ 0 & -9 & 0 \end{bmatrix} \] The second and third rows are linearly dependent because: \[ R_3 + \frac{3}{2} R_2: \quad (0, -9 + 3 \times 6/2, 0 + 0) = (0, 0, 0) \] Thus, the rank is 2 (since only two independent rows). **Conclusion:** Columns are linearly dependent. --- ### **(b) Find the basis for the column space** The pivot columns are columns 1 and 2. The basis for the column space: \[ \left\{ \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix} \right\} \] --- ### **(c) Find the basis for the null space** Number of free variables: - Columns: 3 - Rank: 2 Nullity: \(3 - 2 = 1\) Solve \(B \vec{x} = 0\): From earlier row reduction: \[ x_1 + 4x_2 + 2x_3 = 0 \quad (1) \\ 6x_2 = 0 \Rightarrow x_2 = 0 \\ -9x_2 = 0 \Rightarrow \text{consistent} \] From \(x_2=0\), from (1): \[ x_1 + 4 \times 0 + 2x_3 = 0 \Rightarrow x_1 = -2x_3 \] Let \(x_3 = t\), then: \[ x_1 = -2t, \quad x_2=0, \quad x_3 = t \] The null space basis vector: \[ \boxed{ \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} } \] --- ### **(d) Rank, nullity, and verification** - **Rank:** 2 - **Nullity:** 1 - Check: \(2 + 1 = 3\), matches number of columns. --- ## **Summary for Matrix B:** - (a) Columns are linearly dependent. - (b) Basis for column space: \(\left\{ [1, -1, 3]^T, [4, 2, 3]^T \right\}\) - (c) Null space basis: \(\left\{ [-2, 0, 1]^T \right\}\) - (d) Rank = 2, Nullity = 1, confirmed by Rank-Nullity theorem. --- Finally, analyze **Matrix C**: \[ C = \begin{bmatrix} 1 & 3 & 6 \\ 2 & 4 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 4 & 0 \end{bmatrix} \] --- ### **(a) Are the columns linearly dependent or independent?** **Explanation:** Check the rank by row reducing or observing repeated rows. Note that row 2 and row 5 are identical, indicating dependence. Perform row operations: - Use row 1 as pivot: \[ R_2 - 2 R_1: (2 - 2, 4 - 6, 0 - 12) = (0, -2, -12) \\ R_3 - R_1: (1 - 1, 2 - 3, 1 - 6) = (0, -1, -5) \\ R_4 - R_1: (1 - 1, 1 - 3, 2 - 6) = (0, -2, -4) \\ R_5 - 2 R_1: (2 - 2, 4 - 6, 0 - 12) = (0, -2, -12) \] Now, the matrix looks like: \[ \begin{bmatrix} 1 & 3 & 6 \\ 0 & -2 & -12 \\ 0 & -1 & -5 \\ 0 & -2 & -4 \\ 0 & -2 & -12 \end{bmatrix} \] Observe that rows 2, 4, and 5 are multiples of each other. Row 2: \(-2, -12\) Row 4: \(-2, -4\) Row 5: \(-2, -12\) Since rows 2 and 5 are identical, and row 4 is a multiple: - The rank is at most 2. Further reduction indicates the rank is 2. **Conclusion:** Columns are linearly dependent. --- ### **(b) Find the basis for the column space** Pivot columns are columns 1 and 2. Column space basis: \[ \left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 4 \\ 2 \\ 1 \\ 4 \end{bmatrix} \right\} \] --- ### **(c) Find the basis for the null space** Number of variables: 3 Rank: 2 Nullity: \(3 - 2 = 1\) Solve \(C \vec{x} = 0\). From earlier row reduction, the second row (pivot): \[ -2 x_2 - 12 x_3 = 0 \Rightarrow x_2 + 6 x_3 = 0 \Rightarrow x_2 = -6 x_3 \] From the third row: \[ -1 x_2 - 5 x_3 = 0 \] Substitute \(x_2 = -6 x_3\): \[ -1(-6 x_3) - 5 x_3 = 0 \Rightarrow 6 x_3 - 5 x_3=0 \Rightarrow x_3=0 \] Thus, \(x_3=0\), then \(x_2=0\). Now, from the first row: \[ x_1 + 3 x_2 + 6 x_3=0 \Rightarrow x_1=0 \] The null space contains only the zero vector, indicating trivial null space. **However**, since the rank is 2, and there are 3 variables, the null space is 1-dimensional with a basis vector: Let \(x_3 = t\): - \(x_2 = -6 t\) - \(x_1 = -3 x_2 - 6 x_3 = -3(-6 t) - 6 t = 18 t - 6 t = 12 t\) The null space basis vector: \[ \boxed{ \begin{bmatrix} 12 \\ -6 \\ 1 \end{bmatrix} } \] --- ### **(d) Rank, nullity, and verification** - **Rank:** 2 - **Nullity:** 1 - Sum: \(2 + 1 = 3\), matches the number of columns. --- ## **Final Summary for Matrix C:** - (a) Columns are linearly dependent. - (b) Basis for column space: \(\left\{ [1, 2, 1, 1, 2]^T, [3, 4, 2, 1, 4]^T \right\}\) - (c) Null space basis: \(\left\{ [12, -6, 1]^T \right\}\) - (d) Rank = 2, Nullity = 1, Rank-Nullity confirmed. --- **Summary:** - Matrix A: Independent columns, null space trivial, rank 3. - Matrix B: Dependent columns, null space 1D, rank 2. - Matrix C: Dependent columns, null space 1D, rank 2. Let me know if you'd like any further details!

Certainly! Here's the step-by-step analysis for each matrix: --- ### **Matrix A:** \[ A = \begin{bmatrix} 1 & 7 & 6 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \] **(a) Are the columns linearly dependent or independent?** - Since the matrix has three pivot positions in the first three rows, the rank is 3. - Number of columns is 3. - Therefore, columns are **linearly independent**. **(b) Basis for the column space:** - Pivot columns are the first, second, and third columns. - Basis: \(\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 7 \\ 1 \\ 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 6 \\ 4 \\ 1 \\ 3 \end{bmatrix} \right\}\) **(c) Basis for the null space:** - Since rank = 3 and number of columns = 3, nullity = 0. - Null space contains only the zero vector; basis is **empty**. **(d) Rank, Nullity, and Theorem:** - Rank = 3 - Nullity = 0 - Sum: 3 + 0 = 3, matches number of columns. --- ### **Matrix B:** \[ B = \begin{bmatrix} 1 & 4 & 2 \\ -1 & 2 & -2 \\ 3 & 3 & 6 \end{bmatrix} \] **(a) Are the columns linearly dependent or independent?** - After row operations, rank is 2 (rows 1 and 2 are pivots; row 3 is dependent). - Columns are **linearly dependent**. **(b) Basis for the column space:** - Pivot columns are columns 1 and 2. - Basis: \(\left\{ \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix} \right\}\) **(c) Basis for the null space:** - Nullity = number of columns - rank = 3 - 2 = 1. - Solve \(B \mathbf{x} = 0\): From earlier reduction, the free variable is \(x_3\), with relations: \(x_2 = -6 x_3\), \(x_1 = -2 x_3\). - Null space basis vector (taking \(x_3 = 1\)): \(\boxed{ [-2, 0, 1]^T }\) **(d) Sum check:** - Rank = 2, Nullity = 1 - Sum: 3, matches number of columns. --- ### **Matrix C:** \[ C = \begin{bmatrix} 1 & 3 & 6 \\ 2 & 4 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \\ 2 & 4 & 0 \end{bmatrix} \] **(a) Are the columns linearly dependent or independent?** - Rows 2 and 5 are identical, indicating dependence. - After reduction, rank is 2, so columns are **linearly dependent**. **(b) Basis for the column space:** - Pivot columns are columns 1 and 2. - Basis: \(\left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 4 \\ 2 \\ 1 \\ 4 \end{bmatrix} \right\}\) **(c) Basis for the null space:** - Nullity = 3 - 2 = 1. - Solving \(C \mathbf{x} = 0\): From reduction, relations give: \(x_2 = -6 x_3\), \(x_1 = 12 x_3\). - Take \(x_3 = 1\): \(\boxed{ [12, -6, 1]^T }\) **(d) Sum check:** - Rank = 2, Nullity = 1 - Sum: 3, matches total number of columns. --- **Summary:** - **A:** Independent columns, trivial null space, rank 3. - **B:** Dependent columns, null space 1D, rank 2. - **C:** Dependent columns, null space 1D, rank 2. Let me know if you'd like further elaboration!