Week 8 (lectures 15 and 16): In this question, you will look at solutions to linear systems of equations of the form $A \\boldsymbol{x} = \\boldsymbol{b}$. This will involve matrices in row reduced echelon form, and the row and column ranks of matrices.
\nIf you have not provided an answer to every input gap of a question or part of the question, and you try to submit your answers to the question or part, then you will see the message \"Can not submit answer - check for errors\". In reality your answer has been submitted, but the system is just concerned that you have not submitted an answer to every input gap. For this reason, please ensure that you provide an answer to every input gap in the question or part before submitting. Even if you are unsure of the answer, write down what you think is most likely to be correct; you can always change your answer or retry the question.
\nAs with all questions, there may be parts where you can choose to \"Show steps\". This may give a hint, or it may present sub-parts which will help you to solve that part of the question. Furthermore, remember to always press the \"Show feedback\" button at the end of each part. Sometimes, helpful feedback will be given here, and often it will depend on how correctly you have answered and will link to other parts of the question. Hence, always retry the parts until you obtain full marks, and then look at the feedback again.
Keep in mind that in order to see the feedback for a particular part of a question, you must provide a full (but not necessarily correct) answer to that part. Do not worry though, as you can look at the feedback and then ammend your answer accordingly.
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This is the first matrix of part a, it's RRE form is the identity matrix.
", "definition": "matrix([a,b,c],[u*a,u*b+d,u*c+f*d],[v*a,v*b+w*d,v*c+w*f*d+g])", "group": "Variables for part a", "templateType": "anything", "name": "A1"}, "RRE_A1": {"description": "This is the RRE form of the matrix A1.
", "definition": "matrix([1,0,0],[0,1,0],[0,0,1])", "group": "Variables for part a", "templateType": "anything", "name": "RRE_A1"}, "l": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "l"}, "v": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "v"}, "k": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))*g", "group": "Variables for part a", "templateType": "anything", "name": "k"}, "m": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "m"}, "w": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "w"}, "f": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "f"}, "h": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "h"}, "n": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "n"}, "Ans1": {"description": "This will form part of the answer to the second sub-part of this part of the question.
", "definition": "matrix([(c*k-b*f*k)/(a*g)],[f*k/g],[-k/g],[1])", "group": "Variables for part a", "templateType": "anything", "name": "Ans1"}, "u": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "u"}, "RRE_A3": {"description": "This is the RRE form of the matirx A3.
", "definition": "matrix([1,0,0],[0,1,0],[0,0,1],[0,0,0],[0,0,0])", "group": "Variables for part a", "templateType": "anything", "name": "RRE_A3"}, "j": {"description": "This is a constant that will form part of our matrices in part a.
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", "definition": "random (-5..5 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "d"}, "a": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random (-5..5 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "a"}, "A3": {"description": "This is the third matrix of part a. It is the same as the matrix A1 but has two additional rows at the bottom.
", "definition": "matrix([a,b,c],[u*a,u*b+d,u*c+f*d],[v*a,v*b+w*d,v*c+w*f*d+g],[0,h*d,h*f*d],[0,j*d,j*f*d+g])", "group": "Variables for part a", "templateType": "anything", "name": "A3"}, "q": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "q"}, "r": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "r"}, "b": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random(-3..3 except(0))*a", "group": "Variables for part a", "templateType": "anything", "name": "b"}, "p": {"description": "This is a constant that will be used in our matrix for part e.
", "definition": "random(-5..5 except(0))", "group": "Variables for part e", "templateType": "anything", "name": "p"}, "g": {"description": "This is a constant that will form part of our matrices in part a.
", "definition": "random (-5..5 except(0))", "group": "Variables for part a", "templateType": "anything", "name": "g"}, "RRE_A2": {"description": "This is the RRE form of the matrix A2.
", "definition": "matrix([1,0,0,(b*f*k-c*k)/(a*g)],[0,1,0, -(f*k)/g],[0,0,1, k/g])", "group": "Variables for part a", "templateType": "anything", "name": "RRE_A2"}, "A2": {"description": "The is the second matrix of part a. It is the same as matrix A1 but has an additional column which ensures that the system A2.x=0 has one-dimension of solutions.
", "definition": "matrix([a,b,c,0],[u*a,u*b+d,u*c+f*d,0],[v*a,v*b+w*d,v*c+w*f*d+g,k])", "group": "Variables for part a", "templateType": "anything", "name": "A2"}}, "advice": "", "variable_groups": [{"variables": ["a", "b", "c", "d", "f", "g", "h", "j", "k", "u", "v", "w", "A1", "A2", "A3", "RRE_A1", "RRE_A2", "RRE_A3", "Ans1"], "name": "Variables for part a"}, {"variables": ["l", "m", "n", "p", "q", "r"], "name": "Variables for part e"}], "preamble": {"css": "", "js": ""}, "parts": [{"scripts": {}, "sortAnswers": false, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "type": "gapfill", "prompt": "i) Consider the system of equations described by $A \\boldsymbol{x} = \\boldsymbol{0}$, where
$A = \\var{{A1}}$ , $\\boldsymbol{x} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$, and $\\boldsymbol{0} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$.
What is the reduced row echelon form of $A$? (Note that you can change the size of the matrix in the gap below, if you need to).
$\\textbf{RRE}(A) =$ [[0]] .
Hence, what is the unique solution to the system described by $A \\boldsymbol{x} = \\boldsymbol{0}$?
$\\boldsymbol{x} =$ [[1]] .
ii) Consider again the system described by $A \\boldsymbol{x} = \\boldsymbol{0}$, but this time we let
$A = \\var{{A2}}$ , $\\boldsymbol{x} = \\begin{pmatrix} w \\\\ x \\\\ y \\\\ z \\end{pmatrix}$, and $\\boldsymbol{0} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$.
What is the reduced row echelon form of $A$? (Note that you can change the size of the matrix in the gap below, if you need to).
$\\textbf{RRE}(A) =$ [[2]] .
The row reduced echelon form of $A$ has three leading $1\\text{s}$ and one free variable. Namely, $z$ is the free variable. Letting $z=t$, what are the solutions to the system described by $A \\boldsymbol{x} = \\boldsymbol{0}$?
$\\boldsymbol{x} =$ [[3]] $t$.
iii) Once again, consider the system described by $A \\boldsymbol{x} = \\boldsymbol{0}$, but this time we let
$A = \\var{{A3}}$ , $\\boldsymbol{x} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$, and $\\boldsymbol{0} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$.
What is the reduced row echelon form of $A$? (Note that you can change the size of the matrix in the gap below, if you need to).
$\\textbf{RRE}(A) =$ [[4]] .
Hence, what is the unique solution to the system described by $A \\boldsymbol{x} = \\boldsymbol{0}$?
$\\boldsymbol{x} =$ [[5]] .
If we replace $\\boldsymbol{0}$ with $ \\boldsymbol{b} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 4 \\\\ 5 \\end{pmatrix}$. Would the system still have a solution, and why? (To answer this, you could try to calculate the reduced row echelon form of the augmented matrix $(A|\\boldsymbol{b})$ directly. However, given the top three zeroes in $\\boldsymbol{b}$ this can be done by inspection and using what you have already calculated.)
The system would have [[6]] solution(s) because [[7]] .
one
", "no
", "infinitely many
"]}, {"minMarks": 0, "maxMarks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "matrix": ["1", 0, 0], "type": "1_n_2", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": false, "displayType": "dropdownlist", "distractors": ["", "", ""], "displayColumns": "1", "variableReplacements": [], "shuffleChoices": true, "unitTests": [], "showFeedbackIcon": true, "marks": 0, "choices": ["the last two rows, of the reduced row echeleon form of the augmented matrix, would represent invalid equations.
", "all rows, of the reduced row echeleon form of the augmented matrix, represent valid equations.
", "nothing really changed after the replacement.
"]}]}, {"scripts": {"mark": {"script": "this.markingFeedback = [];\n//This clears the feedback that would normally be given automatically by the built in script. We just want to show our own feedback.\n\n//For the code below: \"this.credit tells us the proportion of the marks that the student got for this part of the question. For example, if the student scored 1 mark out of three, then \"this.credit\" would equal to 1/3.\nif (this.credit < 1){\n this.markingComment('Not quite. Try again and have a look at this feedback section once each gap has been answered correctly as there will be some useful feedback.');\n}else if (this.credit == 1){\n this.markingComment('Good! Feedback for the first bit of part ii: Like you said, if $n=3$ then we have a unique solution, as we can see in part a)i); and if $n > 3$, we get free variables, as we can see in part a)ii), which leads to more than one solution. Feedback for the second bit of part ii: We can see that having more than three rows will only mean that the reduced row eachelon form of $A$ will have rows of zeroes after the third row, as we can see in part a)iii). This means that the first three rows of $\\\\textbf{RRE}(A|\\\\boldsymbol{0})$ say that the entries of $\\\\boldsymbol{x}$ are all zero, and the rows after the third row just say $0=0$. Again, we saw this in part a)iii).');\n}", "order": "after"}}, "stepsPenalty": 0, "sortAnswers": false, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "prompt": "Now, suppose $A$ is an $m \\times n$ real-valued matrix, $\\boldsymbol{x}$ is an $n \\times 1$ vector of unknowns, and $\\boldsymbol{0}$ is an $m \\times 1$ vector with all entries equal to zero.
\nSuppose we are told that the reduced row echelon form of $A$ has three leading $1\\text{s}$.
\ni) What are the smallest values that $m$ and $n$ can take? (Press the \"Show steps\" button below for a hint).
$m \\geq$ [[0]] , and $n \\geq$ [[1]] .
ii) Suppose further that we are told that $m=3$. For what values of $n$ is the homogeneous system $A \\boldsymbol{x} = \\boldsymbol{0}$ unique? (Press the \"Show steps\" button below for a hint).
[[2]] $\\leq n \\leq$ [[3]] .
Suppose instead that we are told that $n=3$. For what values of $m$ is the homogeneous system $A \\boldsymbol{x} = \\boldsymbol{0}$ unique? (Press the \"Show steps\" button below for a hint).
$ m \\geq$ [[4]] .
Once you have completed and submitted this part of the question, please press the show feedback button. (You can still resubmit again).
", "steps": [{"prompt": "Hint for part i: By definition, each leading $1$ must be in a unique row and a unique column.
\nHint for part ii: For the first bit, notice that in parts a)i) and a)ii) we had $m=3$. This should give you a clue as to the number of solutions we get depending on the value of $n$. Similarly, for the second bit, notice that in parts a)i) and a)iii) we had $n=3$. Again, this should give you a clue as to the number of solutions we get depending on the value of $m$.
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i) Suppose we are told that $\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 7 \\end{pmatrix}$.
What is the row rank of $A$? [[0]]
What is the column rank of $A$? [[1]]
What is the row rank of $(A | \\boldsymbol{b})$? [[2]]
What is the column rank of $(A | \\boldsymbol{b})$? [[3]]
ii) Suppose instead that we are told that $\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 7 & 0 \\\\ 0 & 1 & 7 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}$.
\nWhat is the row rank of $A$? [[4]]
What is the column rank of $A$? [[5]]
What is the row rank of $(A | \\boldsymbol{b})$? [[6]]
What is the column rank of $(A | \\boldsymbol{b})$? [[7]]
iii) Suppose instead that we are told that $\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 2 & 0 \\\\ 0 & 1 & 4 & 1 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$.
\nWhat is the row rank of $A$? [[8]]
What is the column rank of $A$? [[9]]
What is the row rank of $(A | \\boldsymbol{b})$? [[10]]
What is the column rank of $(A | \\boldsymbol{b})$? [[11]]
Once you have complete all gaps in this part of the question, please press the \"Show feedback\" button.
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"extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "prompt": "Continuing from part c above, fill in the following table to demonstrate the number of solutions of the system $A \\boldsymbol{x} = \\boldsymbol{b}$, for each given $\\textbf{RRE} (A | \\boldsymbol{b})$.
(Press the \"Show steps\" button below if you require a hint).
[[0]]
", "steps": [{"prompt": "Notational remark: As described in the lectures, for a given matrix $M$ we denote its row rank by $\\rho(M)$.
We can recall the following from section 16.3 of the lecture notes:
You can now use the information above, and your practice in finding the column and row ranks of matrices from part c, to answer the question.
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", "No Solutions
", "Infinitely many solutions
"], "layout": {"expression": "", "type": "all"}, "marks": 0, "shuffleAnswers": false, "scripts": {}, "warningType": "none", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "minAnswers": 0, "displayType": "radiogroup", "maxAnswers": 0, "unitTests": [], "variableReplacements": [], "shuffleChoices": true, "maxMarks": 0, "choices": ["$\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 0 & 1 \\\\ 0 & 1 & 0 & 1 \\\\ 0 & 0 & 1 & 7 \\end{pmatrix}$
", "$\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 7 & 3 \\\\ 0 & 1 & 7 & 4 \\\\ 0 & 0 & 0 & 3 \\end{pmatrix}$
", "$\\textbf{RRE} (A | \\boldsymbol{b}) = \\begin{pmatrix} 1 & 0 & 2 & 0 \\\\ 0 & 1 & 4 & 1 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$
"]}]}, {"scripts": {}, "stepsPenalty": 0, "sortAnswers": false, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true, "prompt": "Now suppose we have the system of equations described by $A \\boldsymbol{x} = \\boldsymbol{b}$, with
$A = \\begin{pmatrix} 1 & \\var{{l}} & \\var{{m}} \\\\ \\var{{n}} & \\simplify{{n}*{l}+1} & \\lambda \\\\ 0 & \\var{{p}} & \\simplify{{q}*{p}}\\end{pmatrix}$ , $\\boldsymbol{x} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ , and $\\boldsymbol{b} = \\begin{pmatrix} 0 \\\\ \\var{{r}} \\\\ \\mu \\end{pmatrix}$ , where $\\lambda$ and $\\mu$ take unknown values in $\\mathbb{R}$.
We want to determine what values of $\\lambda$ and $\\mu$ will give us one, infinitely many, and no solutions. If you require a hint, please press the \"Show steps\" button below.
i) What value must $\\lambda$ not take in order for this system to have a unique solution?
$\\lambda \\neq$ [[0]] .
ii) What values must $\\lambda$ and $\\mu$ take so that this system has an infinite number of solutions?
$\\lambda =$ [[1]] and $\\mu =$ [[2]] .
iii) What value must $\\lambda$ take, and must $\\mu$ not take, so that this system has no solutions?
$\\lambda =$ [[3]] and $\\mu \\neq$ [[4]] .
As we saw in part d (in the \"Show steps\" section), we can determine the number of solutions of the system $A \\boldsymbol{x} = \\boldsymbol{b}$ by calculating $\\textbf{RRE} (A | \\boldsymbol{b})$. We can then easily determine the row and column ranks of $A$ and $(A| \\boldsymbol{b})$, which tells us the number of solutions of the system.
Hence, the first step in this part of the question is to attempt to calculate the row reduced echelon form of $(A | \\boldsymbol{b})$. However, because of the unknown constants $\\lambda$ and $\\mu$, we will only be able to go so far without trying to divide by some term involving $\\lambda$, which could be equal to to zero. Nonetheless, this \"partly reduced\" form of $(A | \\boldsymbol{b})$ will still be enough to determine the row and column ranks of $A$ and $(A| \\boldsymbol{b})$, which will depend on the values of $\\lambda$ and $\\mu$. We can then see what values of $\\lambda$ and $\\mu$ will give one, infinitely many, or zero solutions.
This is the question for week 8 of the MA100 course at the LSE. It looks at material from chapters 15 and 16.
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