Nonhomogeneous systems

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<center>Author: Nadiah Kristensen (nadiah.org)</center>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part a</h2> Consider the system of linear equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= b_1\\ 2x_1 + 4x_2 + 8x_3 + 12x_4 &= b_2\\ 3x_1 + 6x_2 + 7x_3 + 13x_4 &= b_3 \end{align*} $$ </nowiki> Reduce $[A \; \mathbf{b}]$ to $[U \; \mathbf{c}]$, so that $A\mathbf{x} = \mathbf{b}$ becomes a triangular system $U\mathbf{x} = \mathbf{c}$. <hr> <<linkreplace "Reveal hints" t8n>> Where to start The augmented matrix $[A \; \mathbf{b}]$ is <nowiki> $$ \begin{equation} [A \; \mathbf{b}] = \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Now perform a Gaussian elimination. <<linkreplace "Show me the first step" t8n>> First step of Gaussian elimination To get every element under the top-left "1" equal to zero, we perform two row operations: <ol> <li>$R_2 - 2R_1 \to R_2'$</li> <li>$R_3 - 3R_1 \to R_3'$</li> </ol> That gives: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & -2 & -2 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Next step: The next pivot is the "2" in the second row, third column. What operation will make the "-2" below it equal to zero? <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: $U $ and $ \mathbf{c}$ are: (a) $ U = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & -2 & -2 \end{bmatrix} $, $\mathbf{c} = \begin{bmatrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{bmatrix} $ [[Choose (a) -> parta_incorrect]] (b) $ U = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} $, $ \mathbf{c} = \begin{bmatrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{bmatrix} $ [[Choose (b) -> parta_correct]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> Your answer suggests you stopped after you reached the augmented matrix: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & -2 & -2 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> However, this system is not yet fully reduced. Your goal is to create a "staircase" pattern in the coefficient part of the augmented matrix. For example: <nowiki> $$ \begin{equation*} \begin{bmatrix} \blacksquare & * & * & * \\ 0 & \blacksquare & * & * \\ 0 & 0 & \blacksquare & * \\ 0 & 0 & 0 & \blacksquare \end{bmatrix} \quad\text{or}\quad \begin{bmatrix} \blacksquare & * & * & * & * \\ 0 & 0 & \blacksquare & * & * \\ 0 & 0 & 0 & 0 & \blacksquare \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*} $$ </nowiki> where $\blacksquare$ are the pivots (non-zero value), and $* $ represents any value. Notice each pivot moves at least one column to the right but always one row down. In other words, the staircase can have "long steps" (skipping columns) or "short steps" (consecutive columns), but it must always go down and to the right. The pattern you have now is <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} \blacksquare & * & * & * \\ 0 & 0 & \blacksquare & * \\ 0 & 0 & \blacksquare & * \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Which element do you need to turn into a 0 to make a staircase pattern? (a) Row 3, column 4 <<linkreplace "Choose (a)" t8n>> Incorrect. Please try again. <<mathjax>> <</linkreplace>> (b) Row 3, column 3 <<linkreplace "Choose (b)" t8n>> Correct! You want to turn the -2 highlighted in red below into a 0. <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & \color{red} -2 & -2 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Which row operation would you use? (a) $R_3 - R_2 \to R_3'$ <<linkreplace "Choose (a)" t8n>> Incorrect; that would turn the -2 into a -4. <<mathjax>> <</linkreplace>> (b) $R_3 + R_2 \to R_3'$ <<linkreplace "Choose (b)" t8n>> Correct! The resulting augmented matrix is: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Therefore, the final answer is <nowiki> $$ \begin{equation} U = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{ and } \mathbf{c} = \begin{bmatrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{bmatrix} \end{equation} $$ </nowiki> [[Show me the full worked solution on a single page-> parta_solution]] [[CONTINUE to next part -> partb]] <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> For the system of linear equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= b_1\\ 2x_1 + 4x_2 + 8x_3 + 12x_4 &= b_2\\ 3x_1 + 6x_2 + 7x_3 + 13x_4 &= b_3 \end{align*} $$ </nowiki> When we reduce $[A \; \mathbf{b}]$ to $[U \; \mathbf{c}]$, then $A\mathbf{x} = \mathbf{b}$ becomes a triangular system $U\mathbf{x} = \mathbf{c}$, where <nowiki> $$ \begin{equation} U = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{ and } \mathbf{c} = \begin{bmatrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{bmatrix} \end{equation} $$ </nowiki> [[Show me the full worked solution -> parta_solution]] [[CONTINUE to next part -> partb]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part a: full solution</h2> Consider the system of linear equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= b_1\\ 2x_1 + 4x_2 + 8x_3 + 12x_4 &= b_2\\ 3x_1 + 6x_2 + 7x_3 + 13x_4 &= b_3 \end{align*} $$ </nowiki> Reduce $[A \; \mathbf{b}]$ to $[U \; \mathbf{c}]$, so that $A\mathbf{x} = \mathbf{b}$ becomes a triangular system $U\mathbf{x} = \mathbf{c}$. <hr> Solution: Create the augmented matrix: <nowiki> $$ \begin{equation} [A \; \mathbf{b}] = \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Reduce $[A \; \mathbf{b}]$ to $[U \; \mathbf{c}]$ using Gaussian elimination. $R_2 - 2R_1 \to R_2'$, $R_3 - 3R_1 \to R_3'$: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & -2 & -2 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} - 3b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> $R_3 + R_2 \to R_3'$: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> [[CONTINUE to next part -> partb]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part b</h2> The system of linear equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= b_1\\ 2x_1 + 4x_2 + 8x_3 + 12x_4 &= b_2\\ 3x_1 + 6x_2 + 7x_3 + 13x_4 &= b_3 \end{align*} $$ </nowiki> has the augmented matrix in row-echelon form <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Question: Find the condition on $b_1$, $b_2$, and $b_3$ for $A\mathbf{x} = \mathbf{b}$ to have a solution. <hr> <<linkreplace "Reveal hint : When does a system have a solution?" t8n>> Hint For the system to have a solution, any all-zero row in the coefficient part of the augmented matrix (i.e., left-hand side) must also have a zero in the corresponding entry of the augmented column (i.e. the right-hand side). If it does not, the system is inconsistent. <<linkreplace "More details" t8n>> For example, let's say that $b_1$, $b_2$, and $b_3$ were chosen such that the augmented matrix was <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right. \right] \end{equation} $$ </nowiki> The final row of the augmented matrix gives the equation <nowiki> $$ \begin{align} 0 x_1 + 0 x_2 + 0 x_3 + 0 x_4 &= 1 \\ 0 &= 1 \end{align} $$ </nowiki> but $0=1$ is impossible. That means there is no solution to this problem. In order for this problem to have a solution, we must have a final row that gives the equation $ 0 = 0 $. <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: For $A\mathbf{x} = \mathbf{b}$ to have a solution, we require (a) $b_3 = b_2 = b_1 = 0$ [[Choose (a) -> partb_too_strict]] (b) $b_3 + b_2 - 5b_1 = 0$ [[Choose (b) -> partb_correct]] (c) $b_2 - 2b_1 = 0$ [[Choose (c) -> partb_wrong_row]] (d) No condition is needed; the system always has a solution. [[Choose (d) -> partb_no_condition]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> For the system to have a solution, we require <nowiki> $$ \begin{equation} b_3 + b_2 - 5b_1 = 0 \end{equation} $$ </nowiki> The third row of the row-echelon form is <nowiki> $$ \begin{equation} \left[ \begin{matrix} 0 & 0 & 0 & 0 \end{matrix} \;\middle|\; b_3 + b_2 - 5b_1 \right] \end{equation} $$ </nowiki> For the system to be consistent, an all-zero row on the left-hand side must also have a zero on the right-hand side. [[CONTINUE to next part -> partci]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> You're on the right track. For the system to be consistent, we do need a zero in the final row of the augmented column (i.e., the right-hand side). However, $b_1 = b_2 = b_3 = 0$ is more strict than necessary. The right-hand side of the third row is not $b_3$ alone, it is the expression $b_3 + b_2 - 5b_1$. There are many combinations of $b_1$, $b_2$, and $b_3$ (not just all zeros) that make this expression equal to zero. /* For example, $b_1 = 1$, $b_2 = 2$, $b_3 = 3$ gives $b_3 + b_2 - 5b_1 = 3 + 2 - 5 = 0$. */ [[Redo question -> partb]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> The expression $b_2 - 2b_1$ appears in the second row of the augmented column: <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ \color{red}{b_{2} - 2b_{1}} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> But the second row is not an all-zero row on the left. Therefore, it gives a valid equation regardless of what $b_2 - 2b_1$ equals. The condition for consistency comes from the all-zero row. Which row is that, and what expression must equal zero? [[Redo question -> partb]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> Consider what happens when $b_3 + b_2 - 5b_1 = 4$. The augmented matrix becomes <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ 4 \end{matrix} \right. \right] \end{equation} $$ </nowiki> The final row gives the equation <nowiki> $$ \begin{align} 0 x_1 + 0 x_2 + 0 x_3 + 0 x_4 &= 4 \\ 0 &= 4 \end{align} $$ </nowiki> but $0=4$ is impossible. This is an inconsistent system — no solution exists. So a condition on $b_1$, $b_2$, and $b_3$ is needed. What condition would ensure the final row gives $0 = 0$ instead? [[Redo question -> partb]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part c.i</h2> The system $A \mathbf{x} = \mathbf{b}$ with <nowiki> $$ \begin{equation} A = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{bmatrix} \end{equation} $$ </nowiki> has the augmented matrix in row-echelon form <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Question: How many linearly independent columns does $A$ have? Explain. <hr> <<linkreplace "Reveal hints" t8n>> Hints The linearly independent columns are the pivot columns. <<linkreplace "What is a pivot column?" t8n>> When the matrix is in row-echelon form, the pivot in each row is the first non-zero entry in that row. A pivot column is any column that contains a pivot. <<linkreplace "Show me step-by-step" t8n>> Walk along each row from the left and highlight the first non-zero element First row: <nowiki> $$ \begin{equation} \begin{matrix} \rightarrow \\ \\ \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Second row: <nowiki> $$ \begin{equation} \begin{matrix} \\ \rightarrow \\ \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & \color{blue} 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Third row -- no pivot: <nowiki> $$ \begin{equation} \begin{matrix} \\ \\ \rightarrow \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & \color{blue} 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> So the pivot columns are columns 1 and 3. <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: (a) 2 [["Choose (a)" -> partci_no_explanation]] (b) 1 [["Choose (b)" -> partci_incorrect]] (c) 3 [["Choose (c)" -> partci_incorrect]] (d) 4 [["Choose (d)" -> partci_incorrect]] (e) 5 [["Choose (e)" -> partci_incorrect]] (f) 2 because there are 2 pivots. [[Choose (f) -> partci_correct]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> There are two linearly independent columns because there are two pivot columns. Always remember: if the question asks you to "describe" or "explain" something, you will lose marks if you only give the solution and don't write anything more. [[CONTINUE to next part -> partcii]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Incomplete.</h2> You are right that there are 2 linearly independent columns, but if you're asked to explain your answer and don't write anything, you will lose marks. [[Redo question -> partci]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> To identify the linearly independent columns from the augmented matrix, you must identify the pivot columns. A pivot column is any column that contains a pivot. A pivot is the first non-zero entry in a row. To find the pivots, walk along each row from left to right and highlight the first non-zero element. First row: <nowiki> $$ \begin{equation} \begin{matrix} \rightarrow \\ \\ \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Second row: <nowiki> $$ \begin{equation} \begin{matrix} \\ \rightarrow \\ \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & \color{blue} 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> Third row -- no pivot: <nowiki> $$ \begin{equation} \begin{matrix} \\ \\ \rightarrow \\ \end{matrix} \left[ \left. \begin{matrix} \color{blue} 1 & 2 & 3 & 5 \\ 0 & 0 & \color{blue} 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> So the pivot columns are columns 1 and 3. [[Redo question -> partci]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part c.ii</h2> The system $A \mathbf{x} = \mathbf{b}$ with <nowiki> $$ \begin{equation} A = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{bmatrix} \end{equation} $$ </nowiki> has the augmented matrix in row-echelon form <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} \mathbf{1} & 2 & 3 & 5 \\ 0 & 0 & \mathbf{2} & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> where the pivots have been bolded. Question: Describe the columnspace of $A$. <hr> <<linkreplace "Reveal hints" t8n>> Hints To describe a vector space like a columnspace, you describe it in terms of its basis vectors. You write it in the form: <nowiki> $$ \begin{equation} \text{basis}(C(A)) = \left\{ \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}, \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}, \ldots \right\} \end{equation} $$ </nowiki> or the form <nowiki> $$ \begin{equation} C(A) = \text{span} \left\{ \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}, \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}, \ldots \right\} \end{equation} $$ </nowiki> where each vector is a basis vector of $ A$. You can also add a verbal description of the columnspace, e.g., "it is a 2-dimensional space embedded in 4-dimensions" or "it is a plane embedded in $ \mathbb{R}^4$". <<linkreplace "How do I get the basis vectors of the columnspace of A?" t8n>> The basis vectors of the columnspace of matrix $A$ are the pivot columns of $A$. <<mathjax>> <</linkreplace>> <<linkreplace "How do I determine the dimensionality of the columnspace?" t8n>> The dimension of the columnspace is the number of basis vectors. For example, the columnspace below is 2-dimensional (a plane) <nowiki> $$ C(A) = \text{span} \left\{ \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}_{\color{red} 1}, \begin{bmatrix} \cdot \\ \cdot \\ \cdot \end{bmatrix}_{\color{red} 2} \right\} $$ </nowiki> <<mathjax>> <</linkreplace>> <<linkreplace "How do I determine the dimensionality of the embedding space?" t8n>> The dimension of the space the columnspace is embedded in is the number of of components of its basis vectors. For example, the columnspace below is embedded in 3-dimensional space (in $\mathbb{R}^3$) <nowiki> $$ C(A) = \text{span} \left\{ \begin{matrix} _{\color{red} 1} \\ _{\color{red} 2} \\ _{\color{red} 3} \end{matrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix} \right\} $$ </nowiki> <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: The columnspace of $A$ has the basis (a) $\text{basis}(C(A)) = \left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}3\\2\\0\end{bmatrix}\right\}$ [[Choose (a) -> partcii_incorrect]] (b) $\text{basis}(C(A)) = \left\{\begin{bmatrix}1\\2\\3\end{bmatrix}, \begin{bmatrix}3\\8\\7\end{bmatrix}\right\}$ [[Choose (b) -> partcii_correct]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> The columnspace of $A$ has the basis <nowiki> $$ \begin{equation} \text{basis}(C(A)) = \left\{\begin{bmatrix}1\\2\\3\end{bmatrix}, \begin{bmatrix}3\\8\\7\end{bmatrix}\right\} \end{equation} $$ </nowiki> And how many dimensions does the columnspace have? In what dimension is it embedded? (b.1) : The columnspace is 2-dimensional and is embedded in 3 dimensions <<linkreplace "Choose (b.1)" t8n>> Correct! The columnspace is a plane embedded in $ \mathbb{R}^3 $. <nowiki> $$ \begin{align} \text{basis}(C(A)) = & \left\{ \begin{matrix} \color{blue} _1 \\ \color{blue} _2 \\ \color{blue} _3 \end{matrix} \begin{bmatrix}1\\2\\3\end{bmatrix}_{\color{red} 1}, \begin{bmatrix}3\\8\\7\end{bmatrix}_{\color{red} 2} \right\} \qquad {\color{blue} \text{dimension of embedding = 3}} \\ \\ & {\color{red} \text{dimension of columnspace = 2}} \end{align} $$ </nowiki> <iframe src="https://www.desmos.com/3d/nbs3lle5wb" width="100%" style="min-height:700px"></iframe> <center> Figure 1: The basis vectors of the columnspace (red) and the columnspace (grey plane). </center> [[CONTINUE to next part -> partd]] <<mathjax>> <</linkreplace>> (b.2) : The columnspace is 3-dimensional and is embedded in 2 dimensions <<linkreplace "Choose (b.2)" t8n>> Incorrect. To determine the dimension of the columnspace, count the number of basis vectors. <nowiki> $$ \begin{equation} \text{basis}(C(A)) = \left\{ \begin{bmatrix}1\\2\\3\end{bmatrix}_{\color{red} 1}, \begin{bmatrix}3\\8\\7\end{bmatrix}_{\color{red} 2} \right\} \qquad \text{dimension of columnspace = 2} \end{equation} $$ </nowiki> To determine the dimension of the embedding space, count the number of components of the basis vectors. <nowiki> $$ \begin{equation} \text{basis}(C(A)) = \left\{ \begin{matrix} \color{red} _1 \\ \color{red} _2 \\ \color{red} _3 \end{matrix} \begin{bmatrix}1\\2\\3\end{bmatrix} \begin{bmatrix}3\\8\\7\end{bmatrix} \right\} \qquad \text{dimension of embedding space = 3} \end{equation} $$ </nowiki> <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Incorrect.</h2> You were right that the pivot columns of the augmented matrix were columns 1 and 3 <nowiki> $$ \begin{equation} \left[ \left. \begin{matrix} \mathbf{1} & 2 & 3 & 5 \\ 0 & 0 & \mathbf{2} & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{equation} $$ </nowiki> and you were right that we therefore need columns 1 and 3 to describe the basis of the columnspace. However, you must take columns 1 and 3 from the original matrix $A$, not the augmented matrix. The reason why is because the row operations performed to obtain the augmented matrix only preserve the rowspace, they do not preserve the columnspace. [[Redo question -> partcii]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part d</h2> The system $A \mathbf{x} = \mathbf{b}$ with <nowiki> $$ \begin{equation} A = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{bmatrix} \end{equation} $$ </nowiki> has the augmented matrix in row-echelon form <nowiki> $$\left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right]$$ </nowiki> Question: Describe the nullspace of $A$. Find the special solutions in $\mathbb{R}^4$. <hr> <<linkreplace "Reveal hints" t8n>> <<linkreplace "What does 'special solutions' mean?" t8n>> What does 'special solutions' mean? "Special solutions" is just another way of saying "the basis vectors of the nullspace of $A$". They are called special solutions because they can be read off directly using the special-solutions method, without needing to write out the full general solution first. <<mathjax>><</linkreplace>> <<linkreplace "How do I describe the nullspace?" t8n>> How do I describe the nullspace? You can describe the nullspace by finding its basis vectors, i.e., an expression like <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}\cdot \\\cdot \\\cdot \\\cdot \end{bmatrix}, \begin{bmatrix}\cdot \\\cdot \\\cdot\\\cdot\end{bmatrix}, \ldots \right\} \end{equation} $$ </nowiki> This expression says that the nullspace of $A$ is the set of all vectors that can be expressed as a linear combination of the basis vectors. In other words, it is every point that can be reached by a linear combination of the vectors in the basis set. The number of basis vectors equals the number of free variables. To find the basis vectors, solve $A \mathbf{x} = \mathbf{0}$. <<mathjax>> <</linkreplace>> <<linkreplace "Method 1: Show me an overview of how to solve this using the first method we were taught" t8n>> Method 1: First method you were taught Solving $A\mathbf{x} = \mathbf{0}$ gives the equations <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= 0\\ 2x_3 + 2x_4 &= 0 \end{align*} $$ </nowiki> First, identify which are the free variables (there are two of them corresponding to the non-pivot columns). Next, call the free variables $s$ and $t$, write the pivot variables as expressions in terms of $s$ and $t$, and write the solution in the form <nowiki> $$ \begin{equation} \mathbf{x} = \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} = s\begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} + t\begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} \end{equation} $$ </nowiki> Those two vectors are the basis vectors of the nullspace of $A$, also called the special solutions. <<mathjax>> <</linkreplace>> <<linkreplace "Method 2: Show me the first step of the special-solutions method" t8n>> Method 2: Special solutions method Again, start with the equations from the augmented matrix <nowiki> $$ \begin{align} x_1 + 2x_2 + 3x_3 + 5x_4 &= 0 \tag{1} \\ 2x_3 + 2x_4 &= 0 \tag{2} \end{align} $$ </nowiki> Identify the free variables, which correspond to the non-pivot columns. They are $x_2$ and $x_4$. To perform the special solutions method, for each free variable in turn, we set the focal free variable to 1 and all other free variables to 0 and solve. (1) Set $x_2=1$ and $x_4=0$. By back-substitution, Eq. 2 gives $x_3 = 0$, and Eq. 1 gives $x_1 = -2$. Therefore, the first special solution is <nowiki> $$ \begin{bmatrix}-2\\1\\0\\0\end{bmatrix} $$ </nowiki> (2) Set $x_4=1$ and $x_2=0$. Now you do the rest... <<mathjax>> <</linkreplace>> <<mathjax>><</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: The nullspace of $A$ is (a) $ \text{basis}(N(A)) = \left\{\begin{bmatrix}1\\-1/2\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\1\\-1\end{bmatrix}\right\} $ [[Choose (a) -> partd_mixup]] (b) $\text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\-1\\1\end{bmatrix}\right\}$ [[Choose (b) -> partd_x4_only]] (c) $ \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} $ [[Choose (c) -> partd_correct]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> [[Show me the full worked solution -> partd_solution]] [[CONTINUE to next part -> partei]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> You gave the answer <nowiki>$$ \begin{equation} \text{basis}(N(A)) = \left\{ \begin{bmatrix}-2\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\-1\\1\end{bmatrix} \right\} \end{equation} $$ </nowiki> This answer suggests you only used $x_4$ as your free variable, but $x_2$ is also a free variable. Another hint was given in the question: Find the special solutions in $\mathbb{R}^4$. The answer above is only in $\mathbb{R}^3$ (each basis vector has 3 components). [[Redo question -> partd]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> Your answer suggests you mixed up your pivot variables (p) and free variables (f). The free variables are $x_2$ and $x_4$. <nowiki> $$ \begin{align} & \phantom{\bigg[} \left. \begin{matrix} \text{p} & \text{f} & \text{p} & \text{f} \end{matrix} \right. \phantom{\bigg]} \\ & \left[ \left. \begin{matrix} \mathbf{1} & 2 & 3 & 5 \\ 0 & 0 & \mathbf{2} & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right] \end{align} $$ </nowiki> [[Redo question -> partd]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part d: full solution</h2> The system $A \mathbf{x} = \mathbf{b}$ with <nowiki> $$ \begin{equation} A = \begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 4 & 8 & 12 \\ 3 & 6 & 7 & 13 \end{bmatrix} \end{equation} $$ </nowiki> has the augmented matrix in row-echelon form <nowiki> $$\left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right]$$ </nowiki> Question: Describe the nullspace of $A$. Find the special solutions in $\mathbb{R}^4$. <hr> Solution From the augmented matrix we find the pivots <nowiki> $$\left[ \left. \begin{matrix} \mathbf{1} & 2 & 3 & 5 \\ 0 & 0 & \mathbf{2} & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right]$$ </nowiki> There are two pivot variables: $x_1$ and $x_3$. Therefore, we let $x_2 = s$ and $x_4 = t$ be free variables. The augmented matrix gives the equations <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= 0\\ 2x_3 + 2x_4 &= 0 \end{align*} $$ </nowiki> Given $x_4 = t$, Eq. 2 gives: $x_3 = -t$ Substituting into Eq. 1 gives: $x_1 = -2s - 2t$ Therefore <nowiki> $$ \begin{equation} \mathbf{x} = \begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} = \begin{bmatrix}-2s-2t\\s\\-t\\t\end{bmatrix} = s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} \end{equation} $$ </nowiki> What this expression is saying is that the nullspace of $A$ is every point that can be reached by a linear combination of the two vectors $(-2, 1, 0, 0)$ and $(-2, 0, -1, 1)$. Therefore, those two vectors are the basis vectors of the nullspace of $A$ <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} \end{equation} $$ </nowiki> The basis vectors are also known as "special solutions" and they can be obtained using the "special solutions" method. <<linkreplace "Bonus: Show me the special solutions method" t8n>> Special solutions method To perform the special solutions method, for each free variable in turn, we set the focal free variable to 1 and all other free variables to 0 and solve. We have two free variables, $x_2$ and $x_4$. (1) Set $x_2=1$ and $x_4=0$. By back-substitution, that gives $x_3 = 0$, $x_1 = -2$. Therefore, the first special solution is <nowiki> $$ \begin{bmatrix}-2\\1\\0\\0\end{bmatrix} $$ </nowiki> (2) Set $x_4=1$ and $x_2=0$. By back-substitution, that gives $x_3 = -1$, $x_1 = -2$. Therefore, the second special solution is: <nowiki> $$ \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $$ </nowiki> <<mathjax>> <</linkreplace>> [[CONTINUE to next part -> partei]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part e.i: bonus question</h2> Consider a system $A \mathbf{x} = \mathbf{b}$ where $A$ is a matrix with 3 rows and 4 columns. Question: If $ \mathbf{b} $ is the sum of the first and fourth columns of $ A $, what is the particular solution $ \mathbf{x}_P $? <hr> Hints You don't need to do a Gaussian elimination to solve this problem. <<linkreplace "Reveal more hints" t8n>> Remember that $ A \mathbf{x} $ can be written as a linear combination of its columns <nowiki> $$ \begin{equation} A \mathbf{x} = \begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} = x_1 \underbrace{\begin{bmatrix} a_{1,1} \\ a_{2,1} \\ a_{3,1} \end{bmatrix}}_{\text{col. 1}} + x_2 \underbrace{\begin{bmatrix} a_{1,2} \\ a_{2,2} \\ a_{3,2} \end{bmatrix}}_{\text{col. 2}} + x_3 \underbrace{\begin{bmatrix} a_{1,3} \\ a_{2,3} \\ a_{3,3} \end{bmatrix}}_{\text{col. 3}} + x_4 \underbrace{\begin{bmatrix} a_{1,4} \\ a_{2,4} \\ a_{3,4} \end{bmatrix}}_{\text{col. 4}} \end{equation} $$ </nowiki> <<linkreplace "Reveal another hint" t8n>> The question tells you that $ \mathbf{b} $ is the sum of the first and fourth columns of $ A $. So <nowiki> $$ \begin{equation} \overbrace{ x_1 \underbrace{\begin{bmatrix} a_{1,1} \\ a_{2,1} \\ a_{3,1} \end{bmatrix}}_{\text{col. 1}} + x_2 \underbrace{\begin{bmatrix} a_{1,2} \\ a_{2,2} \\ a_{3,2} \end{bmatrix}}_{\text{col. 2}} + x_3 \underbrace{\begin{bmatrix} a_{1,3} \\ a_{2,3} \\ a_{3,3} \end{bmatrix}}_{\text{col. 3}} + x_4 \underbrace{\begin{bmatrix} a_{1,4} \\ a_{2,4} \\ a_{3,4} \end{bmatrix}}_{\text{col. 4}} }^{A \mathbf{x}} = \overbrace{ \underbrace{\begin{bmatrix} a_{1,1} \\ a_{2,1} \\ a_{3,1} \end{bmatrix}}_{\text{col. 1}} + \underbrace{\begin{bmatrix} a_{1,4} \\ a_{2,4} \\ a_{3,4} \end{bmatrix}}_{\text{col. 4}} }^{\mathbf{b}} \end{equation} $$ </nowiki> What are the values of $x_1, x_2, \ldots $? <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: The particular solution $ \mathbf{x}_P $ is (a) $ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} $ <<linkreplace "Choose (a)" t8n>> Correct! [[CONTINUE to next part -> parteii]] <<mathjax>> <</linkreplace>> (b) $ \begin{bmatrix} a_{1,1} + a_{1,4} \\ a_{2,1} + a_{2,4} \\ a_{3,1} + a_{3,4} \\ \end{bmatrix} $ <<linkreplace "Choose (b)" t8n>> That is $ \mathbf{b} $, the right-hand side of the equation. <nowiki> $$ \begin{equation} \underbrace{ \begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} \end{bmatrix} }_{A} \underbrace{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} }_{\mathbf{x}_P} = \underbrace{\begin{bmatrix} a_{1,1} + a_{1,4} \\ a_{2,1} + a_{2,4} \\ a_{3,1} + a_{3,4} \\ \end{bmatrix}}_{\mathbf{b}} \end{equation} $$ </nowiki> What would $ \mathbf{x}_P $ need to be to make this equation true? Check the Hints if you haven't done so already. <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part e.ii</h2> The system $A \mathbf{x} = \mathbf{b}$ has the augmented matrix in row-echelon form $[U \mid \mathbf{c}] $ <nowiki> $$\left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right]$$ </nowiki> and the nullspace <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} \end{equation} $$ </nowiki> Question: Find a particular solution to $A\mathbf{x} = \mathbf{b} = (0, 6, -6)$ and then the complete solution. <hr> <<linkreplace "Reveal hints" t8n>> Hints <<linkreplace "What is a particular solution?" t8n>> A particular solution is any solution to $A \mathbf{x} = \mathbf{b}$. However, as you've learnt from your work with nullspaces, there can sometimes be an infinite number of solutions $\mathbf{x}$ for a given $\mathbf{b}$. In such cases, the easiest particular solution to find is the one where all the free variables equal 0. <<mathjax>> <</linkreplace>> <<linkreplace "What is a complete solution?" t8n>> A "complete" (or "general") solution has the form <nowiki> $$ \begin{equation*} \mathbf{x}_G = \mathbf{x}_P + \mathbf{x}_H \end{equation*} $$ </nowiki> where $\mathbf{x}_P$ is a particular solution and $\mathbf{x}_H$ is the homogeneous solution. <<linkreplace "What is the homogeneous solution?" t8n>> The homogeneous solution $ \mathbf{x}_H $ is the solution to <nowiki> $$ \begin{equation} A \mathbf{x} = \mathbf{0} \end{equation} $$ </nowiki> It has the form <nowiki> $$ \begin{equation} \mathbf{x}_H = k_1 \begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} + k_2 \begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} + k_3 \begin{bmatrix} \cdot \\ \cdot \\ \cdot \\ \cdot \end{bmatrix} + \ldots \end{equation} $$ </nowiki> where the $k_i$ are arbitrary constants and each corresponding vector is a basis vector of the nullspace of $ A $. <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <<mathjax>> <</linkreplace>> <hr> <<linkreplace "Show me step-by-step" t8n>> Guided solution The "complete" (or "general") solution is <nowiki> $$ \begin{equation*} \mathbf{x}_G = \mathbf{x}_P + \mathbf{x}_H \end{equation*} $$ </nowiki> where $\mathbf{x}_P$ is a particular solution and $\mathbf{x}_H$ is the homogeneous solution. We'll find each in turn below. Step 1: Find a particular solution $\mathbf{x}_P$ A particular solution is any solution to $A \mathbf{x} = \mathbf{b}$. In this question, we have a non-trivial nullspace ($N(A) \neq \mathbf{0}$), so we have an infinite number of particular solutions. Therefore, we will find the easiest particular solution, which is when all the free variables equal 0. We have already reduced our system to row-echelon form $[U \mid \mathbf{c}] $, which will give us our system of equations to solve. When we substitute $\mathbf{b} = (0, 6, -6)^T$ into our expression for $\mathbf{c}$ above, we obtain (a) $ \mathbf{c} = \begin{bmatrix} 0 \\ 6 \\ 0 \end{bmatrix} $ <<linkreplace "Choose (a)" t8n>><<replace "#find_c_wrong">><</replace>><<run $("#found_c").show()>>Correct!<</linkreplace>> (b) $ \mathbf{c} = \begin{bmatrix} 0 \\ 6 \\ 12 \end{bmatrix} $ <<linkreplace "Choose (b)" t8n>> Incorrect. When $\mathbf{b} = (0, 6, -6)^T$: <nowiki> $$ \begin{equation*} \mathbf{c} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \\ b_3 + b_2 - 5b_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 - 2(0) \\ -6 + 6 - 5(0) \end{bmatrix} = \ldots ? \end{equation*} $$ </nowiki> <<mathjax>><</linkreplace>> From $U\mathbf{x} = \mathbf{c}$ <nowiki> $$ \begin{equation*} [\mathbf{U} \mid \mathbf{c}] = \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} 0 \\ 6 \\ 0 \end{matrix} \right. \right] \end{equation*} $$ </nowiki> we obtain the simultaneous equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= 0 \quad (1) \\ 2x_3 + 2x_4 &= 6 \quad (2) \\ 0 &= 0 \end{align*} $$ </nowiki> To solve the equations, we will set all the free variables to zero. The free variables are: (a) $x_1$ and $ x_3 $ <<linkreplace "Choose (a)" t8n>> Incorrect; those are the pivot variables. The free variables are the variables corresponding to the non-pivot columns <nowiki> $$ \begin{align} & \phantom{\bigg[} \left. \begin{matrix} \text{p} & \text{f} & \text{p} & \text{f} \end{matrix} \right. \phantom{\bigg]} \\ & \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} 0 \\ 6 \\ 0 \end{matrix} \right. \right] \end{align} $$ </nowiki> <<mathjax>><</linkreplace>> (b) $x_2$ and $ x_4 $ <<linkreplace "Choose (b)" t8n>><<replace "#find_freevar_wrong">><</replace>><<run $("#found_freevar").show()>>Correct!<</linkreplace>> So we set the free variables: <nowiki> $$ \begin{align*} & x_2 = 0 \\ & x_4 = 0 \end{align*} $$ </nowiki> Substituting $ x_2 = x_4 = 0 $ into Eq. 2 above, we obtain (a) $ x_3 = 2 $ <<linkreplace "Choose (a)" t8n>> Incorrect. Remember we have set the free variable $ x_4 = 0 $. <<mathjax>><</linkreplace>> (b) $x_3 = 3$ <<linkreplace "Choose (b)" t8n>><<replace "#find_x3_wrong">><</replace>><<run $("#found_x3").show()>>Correct!<</linkreplace>> <nowiki> $$ \begin{align*} & 2 x_3 + 2(0) = 6 \\ & \implies x_3 = 3 \end{align*} $$ </nowiki> And substituting $ x_2 = x_4 = 0 $ and $x_3 = 3 $ into Eq. 1 above, we obtain <nowiki> $$ \begin{align*} & x_1 + 2(0) + 3(3) + 5(0) = 0 \\ & \implies x_1 = -9 \end{align*} $$ </nowiki> Therefore, the particular solution is (a) $ \mathbf{x}_P = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} $ <<linkreplace "Choose (a)" t8n>><<replace "#find_xp_wrong">><</replace>><<run $("#found_xp").show()>>Correct!<</linkreplace>> (b) $ \mathbf{x}_P = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}-9\\3\\0\\0\end{bmatrix} $ <<linkreplace "Choose (b)" t8n>> Incorrect. You've substituted $x_2$ and $x_3$ incorrectly. <<mathjax>><</linkreplace>> Step 2: Find the homogeneous solution $\mathbf{x}_H$ We were given the nullspace basis vectors in the question: <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} \end{equation} $$ </nowiki> Therefore the homogeneous solution is <nowiki> $$ \mathbf{x}_H = s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $$ </nowiki> The general solution combines the particular and homogeneous solutions: <nowiki> $$ \begin{equation} \mathbf{x}_{G} = \mathbf{x}_P + \mathbf{x}_H = \ldots \quad ? \end{equation} $$ </nowiki> Enter your answer below. <<mathjax>><</linkreplace>> <hr> <<linkreplace "I'm ready to answer" t8n>> Answer: The complete solution is (a) $ \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} $ [[Choose (a) -> parteii_particular_only]] (b) $ s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $ [[Choose (b) -> parteii_homogeneous_only]] (c) $ \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} + s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $ [[Choose (c) -> parteii_correct]] (d) $ \begin{bmatrix}9\\0\\3\\0\end{bmatrix} + s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $ [[Choose (d) -> parteii_sign_error]] <<mathjax>> <</linkreplace>>

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Correct!</h2> The complete solution to $A\mathbf{x} = \mathbf{b} = (0, 6, -6)$ is <nowiki> $$ \begin{equation} \mathbf{x}_{G} = \mathbf{x}_P + \mathbf{x}_H = \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} + s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} \end{equation} $$ </nowiki> where the particular solution $\mathbf{x}_P = (-9, 0, 3, 0)^T$ is found by setting the free variables to zero, and the homogeneous solution $\mathbf{x}_H$ comes from the nullspace basis vectors. [[Show me the full worked solution -> parteii_solution]] [[CONTINUE to next part -> end]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> You found the particular solution <nowiki> $$ \begin{equation} \mathbf{x}_P = \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} \end{equation} $$ </nowiki> which is correct, well done. However, the question asked for the complete solution. The complete (or general) solution has the form <nowiki> $$ \begin{equation} \mathbf{x}_G = \mathbf{x}_P + \mathbf{x}_H \end{equation} $$ </nowiki> where $\mathbf{x}_H$ is the homogeneous solution, which is an an arbitrary linear combination of the nullspace basis vectors. You need to include $\mathbf{x}_H$ because there are infinitely many solutions, and $\mathbf{x}_P$ is just one of them. [[Redo question -> parteii]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> You wrote the homogeneous solution <nowiki> $$ \begin{equation} \mathbf{x}_H = s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} \end{equation} $$ </nowiki> which solves $A\mathbf{x} = \mathbf{0}$. But the question asked for the solution to $A\mathbf{x} = \mathbf{b}$ where $\mathbf{b} = (0, 6, -6)^T$, not $\mathbf{0}$. The complete solution has the form <nowiki> $$ \begin{equation} \mathbf{x}_G = \mathbf{x}_P + \mathbf{x}_H \end{equation} $$ </nowiki> You need to also find a particular solution $\mathbf{x}_P$, which is any single vector that satisfies $A\mathbf{x} = (0, 6, -6)^T $. The easiest way is to set all free variables to zero and solve by back-substitution. [[Redo question -> parteii]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Not quite.</h2> You have the right structure -- particular solution plus homogeneous solution -- but there is an arithmetic error in the particular solution. You wrote $\mathbf{x}_P = (9, 0, 3, 0)^T$, but check the sign of $x_1$. From equation (1) with $x_2 = x_4 = 0$ and $x_3 = 3$: <nowiki> $$ \begin{equation} x_1 + 2(0) + 3(3) + 5(0) = 0 \implies x_1 + 9 = 0 \implies x_1 = -9 \end{equation} $$ </nowiki> [[Redo question -> parteii]]

<center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> <h2>Part e.ii: Full solution</h2> The system $A \mathbf{x} = \mathbf{b}$ has the augmented matrix in row-echelon form $[U \mid \mathbf{c}] $ <nowiki> $$\left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} b_{1} \\ b_{2} - 2b_{1} \\ b_{3} + b_{2} - 5b_{1} \end{matrix} \right. \right]$$ </nowiki> and the nullspace <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} \end{equation} $$ </nowiki> Question: Find a particular solution to $A\mathbf{x} = \mathbf{b} = (0, 6, -6)$ and then the complete solution. <hr> Solution When $\mathbf{b} = (0, 6, -6)^T$: <nowiki> $$ \begin{equation*} \mathbf{c} = \begin{bmatrix} b_1 \\ b_2 - 2b_1 \\ b_3 + b_2 - 5b_1 \end{bmatrix} = \begin{bmatrix} 0 \\ 6 - 2(0) \\ -6 + 6 - 5(0) \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ 0 \end{bmatrix} \end{equation*} $$ </nowiki> We are looking to find a particular solution (just any one solution) of $A\mathbf{x} = \mathbf{b}$, which has the same solutions as $\mathbf{U}\mathbf{x} = \mathbf{c}$. From $U\mathbf{x} = \mathbf{c}$: <nowiki> $$ \begin{equation*} [\mathbf{U} \mid \mathbf{c}] = \left[ \left. \begin{matrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 2 & 2 \\ 0 & 0 & 0 & 0 \end{matrix} \right| \left. \begin{matrix} 0 \\ 6 \\ 0 \end{matrix} \right. \right] \end{equation*} $$ </nowiki> Gives equations: <nowiki> $$ \begin{align*} x_1 + 2x_2 + 3x_3 + 5x_4 &= 0 \quad (1) \\ 2x_3 + 2x_4 &= 6 \quad (2) \\ 0 &= 0 \end{align*} $$ </nowiki> Variables $x_2$ and $x_4$ are free, so we are free to assign them any value in order to find our particular solution. The easiest choice is to set them all to zero. Let $x_2 = x_4 = 0$. Use (2) to to find $x_3$: <nowiki> $$ \begin{equation*} 2x_3 + 2(0) = 6 \Rightarrow x_3 = 3 \end{equation*} $$ </nowiki> Use (1) to find $x_1$: <nowiki> $$ \begin{equation*} x_1 + 2(0) + 3(3) + 5(0) = 0 \Rightarrow x_1 = -9 \end{equation*} $$ </nowiki> Therefore, the particular solution: <nowiki> $$ \begin{equation} \mathbf{x}_P = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} \end{equation} $$ </nowiki> We were given the nullspace basis vectors in the question: <nowiki> $$ \begin{equation} \text{basis}(N(A)) = \left\{\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix}\right\} \end{equation} $$ </nowiki> Therefore the homogeneous solution is <nowiki> $$ \mathbf{x}_H = s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} $$ </nowiki> The general solution combines the particular and homogeneous solutions: <nowiki> $$ \begin{equation} \mathbf{x}_{G} = \mathbf{x}_P + \mathbf{x}_H = \begin{bmatrix}-9\\0\\3\\0\end{bmatrix} + s\begin{bmatrix}-2\\1\\0\\0\end{bmatrix} + t\begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} \end{equation} $$ </nowiki> [[CONTINUE to next part -> end]]

/* <center>(a) -- (b) -- (c.i) -- (c.ii) -- (d) -- (e.i) -- (e.ii) -- (end) </center> */ <h1>Congratulations!</h1> <div style="text-align: center;"> <img src="assets/cat_thinking.jpg" style="width: 300px;"> You have completed the questions. [[Play again -> Start]] <hr> <a href="mailto:nadiah@nadiah.org">Send feedback to: nadiah@nadiah.org</a> </div> <center>Artist credit: Natii Herrera (NatiiHerre)</center>

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