// Numbas version: exam_results_page_options {"name": "MA100 LT Week 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"statement": "

Lent Term Week 2 (lectures 23 and 24): In this question, you will look at linear transformations, the kernal and range of a linear transformation, and changing bases.

Please read the following before attempting the question:

If 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.

As 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.

Furthermore, as with all questions, choosing to reveal the answers will only show you the answers which change every time the question is loaded (i.e. answers to randomised questions); the fixed answers will not be revealed.

", "variables": {"k3": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-d/a , 0 , 0 , 1)", "group": "Variables for part b", "templateType": "anything", "name": "k3"}, "A1": {"description": "

This is the matrix for the transformation T in part b.

", "definition": "matrix([-a*g , -b*g , -c*g , -d*g] , [a*f , b*f , c*f , d*f])", "group": "Variables for part b", "templateType": "anything", "name": "A1"}, "l": {"description": "

This is one of the entries of the first vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "l"}, "A_BB": {"description": "

This is the matrix A with respect to basis B for the domain and basis B for the codomain.

", "definition": "matrix([h , j] , [ii , k])", "group": "Variables for parts c and d", "templateType": "anything", "name": "A_BB"}, "k2": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-c/a , 0 , 1 , 0)", "group": "Variables for part b", "templateType": "anything", "name": "k2"}, "g1": {"description": "

This is the first vector of the basis D.

", "definition": "vector(q , r)", "group": "Variables for parts c and d", "templateType": "anything", "name": "g1"}, "t": {"description": "

This is one of the entries of thesecondvector in the basis D. It is chosen such that g1 and g2 really do form a basis (i.e. are linearly independent).

", "definition": "random(-5..5 except(0) except(s*r/q))", "group": "Variables for parts c and d", "templateType": "anything", "name": "t"}, "A_CD": {"description": "

This is the matrix A with respect to basis C for the domain and basis D for the codomain.

", "definition": "M_BD * matrix([h*l + j*m , h*n + j*p] , [ii*l + k*m , ii*n + k*p])", "group": "Variables for parts c and d", "templateType": "anything", "name": "A_CD"}, "k": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "k"}, "m": {"description": "

This is one of the entries of the first vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "m"}, "s": {"description": "

This is one of the entries of the second vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "s"}, "T_f2_D": {"description": "

This is the vector T(f2) with respect to the basis D.

", "definition": "M_BD * T_f2", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f2_D"}, "ii": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "ii"}, "f": {"description": "

This will appear in the cartesian equation of the range of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "f"}, "h": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "h"}, "f2": {"description": "

This is the second vector of the basis C.

", "definition": "vector(n , p)", "group": "Variables for parts c and d", "templateType": "anything", "name": "f2"}, "n": {"description": "

This is one of the entries of the second vector in the basis C.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "n"}, "j": {"description": "

This will be used in the definition of the transformation T in part c.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "j"}, "k1": {"description": "

This is one of the basis vectors of the kernal of T in part b.

", "definition": "vector(-b/a , 1 , 0 , 0)", "group": "Variables for part b", "templateType": "anything", "name": "k1"}, "c": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "c"}, "d": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "d"}, "a": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "a"}, "T_f1_D": {"description": "

This is the vector T(f1) with respect to the basis D.

", "definition": "M_BD * T_f1", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f1_D"}, "f1": {"description": "

This is the first vector of the basis C.

", "definition": "vector(l , m)", "group": "Variables for parts c and d", "templateType": "anything", "name": "f1"}, "T_f2": {"description": "

This is the vector T(f1).

", "definition": "vector(h*n + j*p , ii*n + k*p)", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f2"}, "M_BD": {"description": "

This is thetransition matrix from basis B to D.

", "definition": "(1/(q*t - r*s))*matrix([t , -s] , [-r , q])", "group": "Variables for parts c and d", "templateType": "anything", "name": "M_BD"}, "q": {"description": "

This is one of the entries of the first vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "q"}, "r": {"description": "

This is one of the entries of the first vector in the basis D.

", "definition": "random(-5..5 except(0))", "group": "Variables for parts c and d", "templateType": "anything", "name": "r"}, "g2": {"description": "

This is the second vector of the basis D.

", "definition": "vector(s , t)", "group": "Variables for parts c and d", "templateType": "anything", "name": "g2"}, "b": {"description": "

This will appear in the cartesian equation of the kernal of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "b"}, "p": {"description": "

This is one of the entries of the second vector in the basis C. It is chosen such that f1 and f2 really do form a basis (i.e. are linearly independent).

", "definition": "random(-5..5 except(0) except(n*m/l))", "group": "Variables for parts c and d", "templateType": "anything", "name": "p"}, "g": {"description": "

This will appear in the cartesian equation of the range of T in part b.

", "definition": "random(-5..5 except(0))", "group": "Variables for part b", "templateType": "anything", "name": "g"}, "r1": {"description": "

This is one of the basis vectors of the range of T in part b.

", "definition": "vector(-g/f , 1 )", "group": "Variables for part b", "templateType": "anything", "name": "r1"}, "T_f1": {"description": "

This is the vector T(f1).

", "definition": "vector(h*l + j*m , ii*l + k*m)", "group": "Variables for parts c and d", "templateType": "anything", "name": "T_f1"}}, "advice": "", "variable_groups": [{"variables": ["a", "b", "c", "d", "f", "g", "k1", "k2", "k3", "r1", "A1"], "name": "Variables for part b"}, {"variables": ["h", "ii", "j", "k", "A_BB", "l", "m", "n", "p", "f1", "f2", "q", "r", "s", "t", "g1", "g2", "T_f1", "T_f2", "M_BD", "T_f1_D", "T_f2_D", "A_CD"], "name": "Variables for parts c and d"}], "preamble": {"css": "", "js": ""}, "parts": [{"scripts": {}, "stepsPenalty": 0, "gaps": [{"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "1", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "1", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "2", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "2", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "3", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "3", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "choices": ["

does

", "

does not

does

", "

does not

"], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "matrix": ["0.2", 0], "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "displayType": "dropdownlist", "minMarks": 0, "showCorrectAnswer": false, "displayColumns": 0, "variableReplacements": [], "shuffleChoices": true, "unitTests": [], "showFeedbackIcon": true, "marks": 0, "type": "1_n_2", "maxMarks": 0}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "1", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "1", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "2", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "2", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "3", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "3", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "choices": ["

does

", "

does not

does

", "

does not

"], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "matrix": ["0.2", 0], "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "displayType": "dropdownlist", "minMarks": 0, "showCorrectAnswer": false, "displayColumns": 0, "variableReplacements": [], "shuffleChoices": true, "unitTests": [], "showFeedbackIcon": true, "marks": 0, "type": "1_n_2", "maxMarks": 0}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "1", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "1", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "2", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "2", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "correctAnswerStyle": "plain", "showCorrectAnswer": false, "allowFractions": false, "notationStyles": ["plain", "en", "si-en"], "mustBeReducedPC": 0, "maxValue": "4", "type": "numberentry", "mustBeReduced": false, "variableReplacements": [], "unitTests": [], "minValue": "4", "marks": "0.2", "correctAnswerFraction": false}, {"scripts": {}, "choices": ["

does

", "

does not

does

", "

does not

"], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "matrix": ["0", "0.2"], "distractors": ["", ""], "variableReplacementStrategy": "originalfirst", "displayType": "dropdownlist", "minMarks": 0, "showCorrectAnswer": false, "displayColumns": 0, "variableReplacements": [], "shuffleChoices": true, "unitTests": [], "showFeedbackIcon": true, "marks": 0, "type": "1_n_2", "maxMarks": 0}], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "gapfill", "showCorrectAnswer": true, "steps": [{"marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "type": "information", "extendBaseMarkingAlgorithm": true, "prompt": "

For parts ii and iii, it may be helpful to recall the following:

Suppose $W$ is a subspace of $\\mathbb{R}^n$, and $W$ has $m$ non-redundant equations in its cartesian description (where $m \\leq n$). Then the dimension of $W$ is $n-m$. That is, each equation is one restriction and it reduces the dimension by $1$.

By \"non-redundant\" we mean that there is no equation which can be written in terms of the other $m-1$ equations. If there were such an equation, say the last one, then it would not be an additional restriction and would not reduce the dimension of $W$ any further.

", "variableReplacementStrategy": "originalfirst", "unitTests": [], "variableReplacements": [], "showFeedbackIcon": true, "showCorrectAnswer": true}], "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "prompt": "

Suppose $T$ is a linear transformation from a finite-dimensional vector space $V$ to a vector space $W$. We recall that the rank-nullity theorem tells us that $\\text{rank} (T) + \\text{nullity} (T) = \\text{dim} (V)$. That is, $\\text{dim} (R(T)) + \\text{dim}(\\text{Ker} (T)) = \\text{dim} (V)$.

In the following, we will be given information about the kernal and range of a possible linear transformation and we must decide, based on the rank-nullity theorem, whether such a linear transformation can really exist. (Press the \"Show steps\" button if you require a hint).

i) Suppose we are told that $T:\\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ has a kernal with basis $\\Bigg\\{ \\begin{pmatrix} 1 \\\\ 1 \\\\ 7 \\end{pmatrix} , \\begin{pmatrix} 3 \\\\ 4 \\\\ 3 \\end{pmatrix} \\Bigg\\}$, and its range has basis $\\Bigg\\{ \\begin{pmatrix} 0 \\\\ 5 \\\\ 1 \\end{pmatrix} \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[0]] , $\\text{dim}(\\text{Ker} (T)) =$ [[1]] , and $\\text{dim} (V) =$ [[2]] . Hence, we can see that the rank-nullity theorem [[3]] hold here, and so the linear transformation $T$ [[4]] really exist.

ii) Suppose we are told that $T:\\mathbb{R}^3 \\longrightarrow \\mathbb{R}^3$ has $\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} \\in \\mathbb{R}^3 \\Big| x_1 + x_2 + 7x_3 = 0 \\Bigg\\}$ and $\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\end{pmatrix} \\in \\mathbb{R}^3 \\Big| 3x_1 + 4x_2 + 3x_3 = 0 \\; \\text{ and } \\; 5x_2 + x_3 = 0 \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[5]] , $\\text{dim}(\\text{Ker} (T)) =$ [[6]] , and $\\text{dim} (V) =$ [[7]] . Hence, we can see that the rank-nullity theorem [[8]] hold here, and so the linear transformation $T$ [[9]] really exist.

iii) Suppose we are told that $T:\\mathbb{R}^4 \\longrightarrow \\mathbb{R}^2$ has $\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} \\in \\mathbb{R}^4 \\Big| 2 x_1 + 4 x_2 + x_4 = 0 \\; \\text{ and } \\; x_1 + 4x_3 = 0 \\Bigg\\}$ and $\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} \\in \\mathbb{R}^2 \\Big| 5x_1 + 2x_2 = 0 \\Bigg\\}$.

In this case, we have $\\text{dim} (R(T)) =$ [[10]] , $\\text{dim}(\\text{Ker} (T)) =$ [[11]] , and $\\text{dim} (V) =$ [[12]] . Hence, we can see that the rank-nullity theorem [[13]] hold here, and so the linear transformation $T$ [[14]] really exist.

", "unitTests": [], "showFeedbackIcon": true, "marks": 0, "sortAnswers": false}, {"scripts": {}, "stepsPenalty": 0, "gaps": [{"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "allowFractions": true, "tolerance": 0, "type": "matrix", "correctAnswer": "k1", "variableReplacements": [], "numRows": "4", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "allowFractions": true, "tolerance": 0, "type": "matrix", "correctAnswer": "k2", "variableReplacements": [], "numRows": "4", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "allowFractions": true, "tolerance": 0, "type": "matrix", "correctAnswer": "k3", "variableReplacements": [], "numRows": "4", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "allowFractions": true, "tolerance": 0, "type": "matrix", "correctAnswer": "r1", "variableReplacements": [], "numRows": "2", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "allowFractions": true, "tolerance": 0, "type": "matrix", "correctAnswer": "A1", "variableReplacements": [], "numRows": "2", "numColumns": "4", "showFeedbackIcon": true, "marks": "3"}], "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "gapfill", "showCorrectAnswer": true, "steps": [{"marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "type": "information", "extendBaseMarkingAlgorithm": true, "prompt": "

Let $\\boldsymbol{c_i}$ respresent the $i^{\\text{th}}$ column of $A$. That is, $A = ( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} )$. Note that we are told in the question that $\\boldsymbol{c_1} = \\begin{pmatrix} \\simplify{-{a}*{g}} \\\\ \\simplify{{a}*{f}} \\end{pmatrix}$.

From part ii we know that the range of $T$ must be equal to $\\text{Lin} (\\boldsymbol{r_1})$.
That is, $A \\boldsymbol{x} \\in \\text{Lin} (\\boldsymbol{r_1})$ for all $\\boldsymbol{x} \\in \\mathbb{R}^4$.
This is equivalent to saying that $x_1 \\boldsymbol{c_1} + x_2 \\boldsymbol{c_2} + x_3 \\boldsymbol{c_3} + x_4 \\boldsymbol{c_4} \\in \\text{Lin} (\\boldsymbol{r_1})$ for all $x_1 , x_2 , x_3 , x_4 \\in \\mathbb{R}$.
Hence, we can see that all $\\boldsymbol{c_1} , \\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ must all be multiples of $\\boldsymbol{r_1}$. Notice that the given value of $\\boldsymbol{c_1}$ is such a multiple.
This helps, but we must still use the rest of the information we are given to determine $A$ explicitely.

From part i we know that the kernal of $A$ is $\\text{Lin}\\{ \\boldsymbol{k_1} ,\\boldsymbol{k_2} ,\\boldsymbol{k_3} \\}$.
You should prove to yourself that this is equivalent to saying that $A \\boldsymbol{k_1} = \\boldsymbol{0}$ , $A \\boldsymbol{k_2} = \\boldsymbol{0}$ , and $A \\boldsymbol{k_3} = \\boldsymbol{0}$.
That is, it is equivalent to saying that $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_1} = \\boldsymbol{0}$ , $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_2} = \\boldsymbol{0}$ , and $( \\boldsymbol{c_1} \\; \\boldsymbol{c_2} \\; \\boldsymbol{c_3} \\; \\boldsymbol{c_4} ) \\boldsymbol{k_3} = \\boldsymbol{0}$.
By using the explicit values of $\\boldsymbol{k_1} ,\\boldsymbol{k_2} ,\\boldsymbol{k_3}$ that we found in part i, and expanding the equations above, we can obtain equations which relate $\\boldsymbol{c_1} , \\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ to each other. Since we already know $\\boldsymbol{c_1}$, we can obtain $\\boldsymbol{c_2} , \\boldsymbol{c_3} , \\boldsymbol{c_4}$ from these equations.

Suppose we are told that a linear transformation $T: \\mathbb{R}^4 \\longrightarrow \\mathbb{R}^2$ has

$\\text{ker} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} \\in \\mathbb{R}^4 \\Big| \\simplify[!noLeadingMinus]{{a}*x_1 + {b}*x_2 + {c}*x_3 + {d}*x_4} = 0 \\Bigg\\}$

and

$\\text{R} (T) = \\Bigg\\{ \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} \\in \\mathbb{R}^2 \\Big| \\simplify[!noLeadingMinus]{{f}*x_1 + {g}*x_2} = 0 \\Bigg\\}$.

i) Consider the cartesian equation of $\\text{ker} (T)$. By letting $x_2 = r$ , $x_3 = s$ , and $x_4 = t$, we can see that the parametric description of $\\text{ker} (T)$ is given by

$\\begin{pmatrix} x_1 \\\\ x_2 \\\\ x_3 \\\\ x_4 \\end{pmatrix} =$ [[0]] $r +$ [[1]] $s +$ [[2]] $t$.

Hence, $B_1 = \\{ \\boldsymbol{k_1} , \\boldsymbol{k_2} , \\boldsymbol{k_3} \\} = \\Bigg\\{$ [[0]] , [[1]] , [[2]] $\\Bigg\\}$ is a basis for $\\text{ker} (T)$.

ii) Now consider the cartesian equation of $\\text{R}(T)$. By letting $x_2 = u$, we can see that the parametric description of $\\text{R}(T)$ is given by

$\\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} =$ [[3]] $u$.

Hence, $B_2 = \\{\\boldsymbol{r_1} \\} = \\Bigg\\{$ [[3]]$\\Bigg\\}$ is a basis for $\\text{R} (T)$.

ii) Suppose further that there is a $2 \\times 4$ real matrix $A$ such that $T (\\boldsymbol{x}) = A \\boldsymbol{x}$ for all $\\boldsymbol{x} \\in \\mathbb{R}^4$, and such that the first column of $A$ is $\\begin{pmatrix} \\simplify{-{a}*{g}} \\\\ \\simplify{{a}*{f}} \\end{pmatrix}$. Use what we have learnt from parts i and ii to determine $A$ explicitely. (If you require a hint then please press the \"Show steps\" button below).

$A=$ [[4]] .

Recall that $A_T$ is equal to the matrix $(T ( \\boldsymbol{e_1}) \\; T(\\boldsymbol{e_2}))$, where $T(\\boldsymbol{e_1})$ and $T(\\boldsymbol{e_2})$ are in vector form.

Consider the linear transformation $T: \\mathbb{R}^2 \\longrightarrow \\mathbb{R}^2$ defined by its effect on the standard basis $B = \\{ \\boldsymbol{e_1} , \\boldsymbol{e_2} \\} = \\bigg\\{ \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} , \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\bigg\\}$:

$T(\\boldsymbol{e_1}) = (\\var{h}) \\boldsymbol{e_1} + (\\var{ii}) \\boldsymbol{e_2}$
$T(\\boldsymbol{e_2}) = (\\var{j}) \\boldsymbol{e_1} + (\\var{k}) \\boldsymbol{e_2}$

Find the matrix $A_T$ representing the transformation $T$. that is, find the matrix $A_T^{B\\rightarrow B}$. (Press the \"Show steps\" button if you require a hint).

$A_T =$ [[0]] .

We recall that the matrix $A_T^{C \\rightarrow D}$ is given by $A_T^{C \\rightarrow D} = \\Big( \\big( T(\\boldsymbol{f_1}) \\big)_D \\; , \\; \\big( T(\\boldsymbol{f_2}) \\big)_D \\Big)$, where $\\big( T(\\boldsymbol{f_1}) \\big)_D$ and $\\big( T(\\boldsymbol{f_2}) \\big)_D$ are the vectors $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ represented in terms of the basis $D$, respectively.

We should first find $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ with respect to the standard basis, and then with respect to the basis $D$. From what we are told in part c, we have that

$T(\\boldsymbol{f_1}) = T((\\var{{l}}) \\boldsymbol{e_1} + (\\var{{m}}) \\boldsymbol{e_2}) =$

", "allowFractions": false, "tolerance": 0, "correctAnswer": "T_f1", "variableReplacements": [], "numRows": "2", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "matrix", "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": false, "prompt": "

$T(\\boldsymbol{f_2}) =T((\\var{{n}}) \\boldsymbol{e_1} + (\\var{{p}}) \\boldsymbol{e_2}) =$

", "allowFractions": false, "tolerance": 0, "correctAnswer": "T_f2", "variableReplacements": [], "numRows": "2", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "matrix", "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "prompt": "

Now we wish to express $T(\\boldsymbol{f_1})$ and $T(\\boldsymbol{f_1})$ with respect to the basis $D$. That is, we wish to find $\\lambda_1 , \\lambda_2$ and $\\mu_1 , \\mu_2$ such that $T(\\boldsymbol{f_1}) = \\lambda_1 \\boldsymbol{g_1} + \\lambda_2 \\boldsymbol{g_2}$ and $T(\\boldsymbol{f_2}) =\\mu_1 \\boldsymbol{g_1} +\\mu_2 \\boldsymbol{g_2}$. Then, we would have that $\\big( T(\\boldsymbol{f_1}) \\big)_D = \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}_D$ and $\\big( T(\\boldsymbol{f_2}) \\big)_D = \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}_D$.

Notice that we can express the equation $T(\\boldsymbol{f_1}) =\\lambda_1\\boldsymbol{g_1} +\\lambda_2\\boldsymbol{g_2}$ as $T(\\boldsymbol{f_1}) = M \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}$, and the equation $T(\\boldsymbol{f_2}) =\\mu_1\\boldsymbol{g_1} +\\mu_2\\boldsymbol{g_2}$ as $T(\\boldsymbol{f_2}) = M \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}$, where $M = ( \\boldsymbol{g_1} , \\boldsymbol{g_2})$.

If we calculate $M^{-1}$ then it will be easy for us to determine $\\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}$ and $\\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}$.

$M^{-1} =$

", "allowFractions": true, "tolerance": 0, "correctAnswer": "M_BD", "variableReplacements": [], "numRows": "2", "numColumns": "2", "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "matrix", "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "prompt": "

Hence, we have that $\\big( T(\\boldsymbol{f_1}) \\big)_D = \\begin{pmatrix} \\lambda_1 \\\\ \\lambda_2 \\end{pmatrix}_D = M^{-1} T(\\boldsymbol{f_1}) =$

", "allowFractions": true, "tolerance": 0, "correctAnswer": "T_f1_D", "variableReplacements": [], "numRows": "2", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "matrix", "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "prompt": "

and $\\big( T(\\boldsymbol{f_2}) \\big)_D = \\begin{pmatrix} \\mu_1 \\\\ \\mu_2 \\end{pmatrix}_D = M^{-1}T(\\boldsymbol{f_2}) =$

", "allowFractions": true, "tolerance": 0, "correctAnswer": "T_f2_D", "variableReplacements": [], "numRows": "2", "numColumns": 1, "showFeedbackIcon": true, "marks": 1}, {"markPerCell": false, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "matrix", "allowResize": false, "variableReplacementStrategy": "originalfirst", "unitTests": [], "showCorrectAnswer": true, "correctAnswerFractions": true, "prompt": "

Hence, we have that $A_T^{C \\rightarrow D} = \\Big( \\big( T(\\boldsymbol{f_1}) \\big)_D \\; , \\; \\big(T(\\boldsymbol{f_2}) \\big)_D \\Big) =$

", "allowFractions": true, "tolerance": 0, "correctAnswer": "A_CD", "variableReplacements": [], "numRows": "2", "numColumns": "2", "showFeedbackIcon": true, "marks": "2"}], "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "prompt": "

Now consider the bases $C = \\{ \\boldsymbol{f_1} , \\boldsymbol{f_2} \\} = \\bigg\\{ \\begin{pmatrix} \\var{{l}} \\\\ \\var{{m}} \\end{pmatrix} , \\begin{pmatrix} \\var{{n}} \\\\ \\var{{p}} \\end{pmatrix} \\bigg\\}$ and $D = \\{ \\boldsymbol{g_1} , \\boldsymbol{g_2} \\} = \\bigg\\{ \\begin{pmatrix} \\var{{q}} \\\\ \\var{{r}} \\end{pmatrix} , \\begin{pmatrix} \\var{{s}} \\\\ \\var{{t}} \\end{pmatrix} \\bigg\\}$.

Find the matrix $A_T^{C \\rightarrow D}$ representing the same transformation $T$, but with respect to the non-standard basis $C$ for the domain of $T$ and the non-standard basis $D$ for the codomain of $T$. (Press the \"Show steps\" button for a breakdown of this question into steps).

$A_T^{C \\rightarrow D} =$ [[0]] .

", "unitTests": [], "showFeedbackIcon": true, "marks": 0, "sortAnswers": false}], "rulesets": {}, "ungrouped_variables": [], "metadata": {"description": "

This is the question for Lent Term week 2 of the MA100 course at the LSE. It looks at material from chapters 23 and 24.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "functions": {}, "tags": [], "extensions": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "name": "MA100 LT Week 2", "type": "question", "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}]}], "contributors": [{"name": "Michael Yiasemides", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1440/"}]}