// Numbas version: finer_feedback_settings {"name": "Matrices: Scalar Multiple (Instructional)", "metadata": {"description": "Scalar Multiplication of matrices, plus used in combination with addition & subtraction", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", ""], "variable_overrides": [[], [], []], "questions": [{"name": "Matrices: Scalar Multiple 01", "extensions": [], "custom_part_types": [], "resources": [["question-resources/matrix_add.gif", "/srv/numbas/media/question-resources/matrix_add.gif"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}], "tags": [], "metadata": {"description": "Scalar Multiplication (pre-defined sizes in answers)", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "

In general if $A$ is an $m \\times n$ matrix with typical element $a_{ij}$ then the product of a number $k$ with $A$ is written $kA$ and has the corresponding elements $ka_{ij}$ .

If $ A= \\left( \\begin{array}{ccc} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\\\ a_{31} & a_{32} \\end{array} \\right) $ then $ kA= \\left( \\begin{array}{ccc} k \\times a_{11} & k \\times a_{12} \\\\ k\\times a_{21} &k\\times a_{22} \\\\ k\\times a_{31} &k\\times a_{32} \\end{array} \\right) $

This operation is called scalar multiplication, but its result is not named \"scalar product\" to avoid confusion, since \"scalar product\" is often used as a synonym for the \"dot product\" or \"inner product\".

", "advice": "

We are asked to carry out scalar multiplications of the given matrices.

This is simply achieved by multiplying each element by the constant:

Using this technique results in:

If $A=\\var{A}$ then: $\\var{k1} A=\\left( \\begin{array}{ccc} \\var{k1} \\times \\var{A[0][0]} & \\var{k1} \\times \\var{A[0][1]}& \\var{k1} \\times \\var{A[0][2]}\\\\ \\var{k1} \\times \\var{A[1][0]} & \\var{k1} \\times \\var{A[1][1]}& \\var{k1} \\times \\var{A[1][2]}\\\\ \\var{k1} \\times \\var{A[2][0]} & \\var{k1} \\times \\var{A[2][1]} & \\var{k1} \\times \\var{A[2][2]} \\end{array} \\right) = \\var{ad1}$

Following the same process gives:

If $B=\\var{B}$ then:

$\\var{k2}B=\\var{ad2}$

If $C=\\var{C}$ then:

$\\var{k3}C=\\var{ad3}$

If $D=\\var{D}$ then:

$\\var{k4}D=\\var{ad4}$

If $E=\\var{EE}$ then:

$\\var{k5}E=\\var{ad5}$

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Carry out the scalar multiplication of the following matrices:

You will need to define the size of the matrix before entering your answer.

If $A=\\var{A}$ then:

$\\var{k1} A=$ [[0]]

If $B=\\var{B}$ then:

$\\var{k2}B=$ [[1]]

If $C=\\var{C}$ then:

$\\var{k3}C=$ [[2]]

If $D=\\var{D}$ then:

$\\var{k4}D=$ [[3]]

If $E=\\var{EE}$ then:

$\\var{k5}E=$ [[4]]

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In general if $A$ is an $m \\times n$ matrix with typical element $a_{ij}$ then the product of a number $k$ with $A$ is written $kA$ and has the corresponding elements $ka_{ij}$ .

", "advice": "

We are asked to carry out scalar multiplications of the given matrices.

This is simply achieved by multiplying each element by the constant:

Using this technique results in:

Following the same process gives:

If $B=\\var{B}$ then:

$\\var{k2}B=\\var{ad2}$

If $C=\\var{C}$ then:

$\\var{k3}C=\\var{ad3}$

If $D=\\var{D}$ then:

$\\var{k4}D=\\var{ad4}$

If $E=\\var{EE}$ then:

$\\var{k5}E=\\var{ad5}$

Carry out the scalar multiplication of the following matrices:

You will need to define the size of the matrix before entering your answer.

If $A=\\var{A}$ then:

$\\var{k1} A=$ [[0]]

If $B=\\var{B}$ then:

$\\var{k2}B=$ [[1]]

If $C=\\var{C}$ then:

$\\var{k3}C=$ [[2]]

If $D=\\var{D}$ then:

$\\var{k4}D=$ [[3]]

If $E=\\var{EE}$ then:

$\\var{k5}E=$ [[4]]

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Matrix calculations can often include a combination of operations such as addition, subtraction and scalar multiplication.

", "advice": "

We are asked to carry out combinations of matrix operations:

If $A=\\var{A}$ and $B=\\var{A2}$ then:

$\\var{k1} A+B=\\var{k1} \\var{A}+\\var{A2}$

$\\var{k1} A+B=\\var{ad1}+\\var{A2}$

$\\var{k1} A+B=\\var{ad1a}$

If $C=\\var{B}$ and $D=\\var{B2}$ then:

$C-\\var{k2}D=\\var{B}-\\var{k2}\\var{B2}$

$C-\\var{k2}D=\\var{B}-\\var{ad2}$

$C-\\var{k2}D=\\var{ad2a}$

If $E=\\var{C}$ and $F=\\var{C2}$ then:

$\\var{k3}E-\\var{k1}F=\\var{k3}\\var{C}-\\var{k1}\\var{C2}$

$\\var{k3}E-\\var{k1}F=\\var{ad3}-\\var{ad3a}$

$\\var{k3}E-\\var{k1}F=\\var{ad3b}$

If $G=\\var{D}$ and $H=\\var{D2}$ then:

$\\var{k4}G+H=\\var{k4}\\var{D}+\\var{D2}$

$\\var{k4}G+H=\\var{ad4}+\\var{D2}$

$\\var{k4}G+H=\\var{ad4a}$

If $I=\\var{EE}$ and $J=\\var{EE2}$ then:

$\\var{k5}I-J=\\var{k5}\\var{EE}-\\var{EE2}$

$\\var{k5}I-J=\\var{ad5}-\\var{EE2}$

$\\var{k5}I-J=\\var{ad5a}$

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"ad3a", "ad3b", "ad4", "ad4a", "ad5", "ad5a"], "variable_groups": [{"name": "A", "variables": ["n1", "m1", "A", "A2"]}, {"name": "B", "variables": ["n2", "m2", "B", "B2"]}, {"name": "C", "variables": ["n3", "m3", "C", "C2"]}, {"name": "D", "variables": ["n4", "m4", "D", "D2"]}, {"name": "E", "variables": ["n5", "m5", "EE", "EE2"]}, {"name": "scalar constants", "variables": ["k1", "k2", "k3", "k4", "k5"]}], "functions": {"calculateTrace": {"parameters": [], "type": "number", "language": "javascript", "definition": ""}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": false, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Carry out the following matrix calculations:

If $A=\\var{A}$ and $B=\\var{A2}$ then:

$\\var{k1} A+B=$ [[0]]

If $C=\\var{B}$ and $D=\\var{B2}$ then:

$C-\\var{k2}D=$ [[1]]

If $E=\\var{C}$ and $F=\\var{C2}$ then:

$\\var{k3}E-\\var{k1}F=$ [[2]]

If $G=\\var{D}$ and $H=\\var{D2}$ then:

$\\var{k4}G+H=$ [[3]]

If $I=\\var{EE}$ and $J=\\var{EE2}$ then:

$\\var{k5}I-J=$ [[4]]

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