7. Let $$ A=\left[\begin{array}{ccc}-4 & 1 & -9 \ 5 & -3 & 12 \ 1 & -1 & 4\end{array} ight] $$ and $$ \mathbf{b}=\left[\begin{array}{l}b_{1} \ b_{2} \ b_{3}\end{array} ight] $$. (a) Transform the augmented matrix of the system $$ A \mathbf{x}=\mathbf{b} $$ to row echelon f (b) Use this to determine whether $$ A \mathbf{x}=\mathbf{b} $$ is consistent for all possible $$ b_{1}, b_{2} $$ whether the columns of $$ A $$ span $$ \left.\mathbb{R}^{3} ight) $$. Give a clear explanation.

Question

7. Let $$ A=\left[\begin{array}{ccc}-4 & 1 & -9 \ 5 & -3 & 12 \ 1 & -1 & 4\end{array}ight] $$ and $$ \mathbf{b}=\left[\begin{array}{l}b_{1} \ b_{2} \ b_{3}\end{array}ight] $$. (a) Transform the augmented matrix of the system $$ A \mathbf{x}=\mathbf{b} $$ to row echelon f (b) Use this to determine whether $$ A \mathbf{x}=\mathbf{b} $$ is consistent for all possible $$ b_{1}, b_{2} $$ whether the columns of $$ A $$ span $$ \left.\mathbb{R}^{3}ight) $$. Give a clear explanation.

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the prompt.

### Part (a): Transform the augmented matrix to row echelon form

1. **Form the augmented matrix**: The augmented matrix for the system $$ A \mathbf{x} = \mathbf{b} $$ is given by:
   $$
   \left[\begin{array}{ccc|c}
   -4 & 1 & -9 & b_1 \
   5 & -3 & 12 & b_2 \
   1 & -1 & 4 & b_3
   \end{array}\right]
   $$

2. **Row operations**: We will perform row operations to transform this matrix into row echelon form.

- **Step 1**: We can start by making the first entry of the first column a leading 1. We can do this by swapping rows or scaling. Here, we can scale Row 3 by 1 to make it easier to work with:
     $$
     R_3 \leftarrow R_3 \cdot 1
     $$
     The matrix becomes:
     $$
     \left[\begin{array}{ccc|c}
     -4 & 1 & -9 & b_1 \
     5 & -3 & 12 & b_2 \
     1 & -1 & 4 & b_3
     \end{array}\right]
     $$

- **Step 2**: Next, we can eliminate the first column entries below the leading entry in Row 3. We can perform the following operations:
     $$
     R_1 \leftarrow R_1 + 4R_3
     $$
     $$
     R_2 \leftarrow R_2 - 5R_3
     $$
     This gives us:
     $$
     \left[\begin{array}{ccc|c}
     0 & -3 & -1 & b_1 + 4b_3 \
     0 & 2 & 8 & b_2 - 5b_3 \
     1 & -1 & 4 & b_3
     \end{array}\right]
     $$

- **Step 3**: Now we can scale Row 2 to make the leading coefficient 1:
     $$
     R_2 \leftarrow R_2 \cdot \frac{1}{2}
     $$
     The matrix becomes:
     $$
     \left[\begin{array}{ccc|c}
     0 & -3 & -1 & b_1 + 4b_3 \
     0 & 1 & 4 & \frac{b_2 - 5b_3}{2} \
     1 & -1 & 4 & b_3
     \end{array}\right]
     $$

- **Step 4**: Finally, we can eliminate the second column entry in Row 1:
     $$
     R_1 \leftarrow R_1 + 3R_2
     $$
     This results in:
     $$
     \left[\begin{array}{ccc|c}
     0 & 0 & 11 & b_1 + 4b_3 + 3\left(\frac{b_2 - 5b_3}{2}\right) \
     0 & 1 & 4 & \frac{b_2 - 5b_3}{2} \
     1 & -1 & 4 & b_3
     \end{array}\right]
     $$

Now we have the augmented matrix in row echelon form.

### Part (b): Determine consistency and span

To determine whether the system $$ A \mathbf{x} = \mathbf{b} $$ is consistent for all possible $$ b_1, b_2, b_3 $$, we need to analyze the row echelon form we obtained.

1. **Consistency**: The system is consistent if there are no rows of the form $$ [0 \, 0 \, 0 | c] $$ where $$ c \neq 0 $$. In our row echelon form, we have:
   $$
   \left[\begin{array}{ccc|c}
   0 & 0 & 11 & \text{(some expression)} \
   0 & 1 & 4 & \text{(some expression)} \
   1 & -1 & 4 & b_3
   \end{array}\right]
   $$
   The first row indicates that $$ 11z = \text{(some expression)} $$, which can always be solved for $$ z $$ as long as the right-hand side is consistent. Thus, the system is consistent for all $$ b_1, b_2, b_3 $$.

2. **Span of columns of $$ A $$**: The columns of $$ A $$ span $$ \mathbb{R}^3 $$ if the rank of $$ A $$ is equal to 3. Since we have three rows and the leading coefficients in the row echelon form indicate that we have three pivot positions, the rank of $$ A $$ is 3. Therefore, the columns of $$ A $$ span $$ \mathbb{R}^3 $$.

### Conclusion

- The system $$ A \mathbf{x} = \mathbf{b} $$ is consistent for all possible $$ b_1, b_2, b_3 $$.
- The columns of $$ A $$ span $$ \mathbb{R}^3 $$.
The system $$ A \mathbf{x} = \mathbf{b} $$ is consistent for all possible $$ b_1, b_2, b_3 $$, and the columns of $$ A $$ span $$ \mathbb{R}^3 $$.

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