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Set of matrices exercises for knowledge validation. Topics covered:

Matrix Addition and Subtraction

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Exercises covering matrix addition and subtraction

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Matrix Addition and Subtraction

Perform the following matrix additions and subtractions:

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To add or subtract matrices, the matrices need to be of the same order (same number of rows and columns). We then add or subtract each corresponding element.

e.g.

\\[ \\begin{pmatrix} a_{1,1} & a_{1,2} \\\\ a_{2,1} & a_{2,2} \\\\ \\end{pmatrix} + \\begin{pmatrix} b_{1,1} & b_{1,2} \\\\ b_{2,1} & b_{2,2} \\\\ \\end{pmatrix} = \\begin{pmatrix} a_{1,1} + b_{1,1} & a_{1,2} + b_{1,2} \\\\ a_{2,1} + b_{2,1} & a_{2,2} + b_{2,2} \\\\ \\end{pmatrix} \\]

The resulting matrix will also be of the same order.

e.g. Part a)

\\[ \\simplify{{maP1a}} + \\simplify{{maP1b}} = \\begin{pmatrix} \\var{maP1a[0][0]} + \\var{maP1b[0][0]} & \\var{maP1a[0][1]} + \\var{maP1b[0][1]} \\\\ \\var{maP1a[1][0]} + \\var{maP1b[1][0]} & \\var{maP1a[1][1]} + \\var{maP1b[1][1]} \\\\ \\end{pmatrix} = \\simplify{{maP1a + maP1b}} \\]

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$\\simplify{{maP1a}}+\\simplify{{maP1b}} =$ [[0]]

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$\\simplify{{maP2a}}+\\simplify{{maP2b}} =$ [[0]]

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$\\simplify{{maP3a}}-\\simplify{{maP3b}} =$ [[0]]

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$\\simplify{{maP4a}}+\\simplify{{maP4b}} =$ [[0]]

$\\simplify{{maP5a}}-\\simplify{{maP5b}} =$ [[0]]

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Matrix by scalar and matrix by matrix multiplication exercises.

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Matrix Multiplication

Answer the following questions on matrix multiplication.

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Scalar Multiplication

For scalar multiplication, multiply each element by the scalar value. The resultant matrix will be of the same order (size) as the starting matrix.

e.g. for the first question:

$\\simplify{{maS1}}\\times\\var{rand1} = \\begin{pmatrix}
\\var{maS1[0][0]}\\times\\var{rand1} & \\var{maS1[0][1]}\\times\\var{rand1} \\\\
\\var{maS1[1][0]}\\times\\var{rand1} & \\var{maS1[1][1]}\\times\\var{rand1}
\\end{pmatrix} = \\simplify{{maS1a}}$

Matrix by Matrix Multiplcation

Matrices can only be multiplied together when the number of columns in the first matrix is equal to the number of rows in the second matrix. The resultant matrix will have the same number of rows as the first matrix, and the same number of columns as the second matrix.

For $AC$,

$A = \\simplify{{maMa}} C = \\simplify{{maMc}}$

The order of $A$ is [3 x 2] and the order of $C$ is [2 x 4].

[3 x 2] x [2 x 4] : The inner numbers match, therefore these matrices can be multiplied. The outer numbers give the size of the new matrix, which will be [3 x 4].

The answer matrix will look like this: $\\begin{pmatrix}
ac_{11} & ac_{12} & ac_{13} & ac_{14} \\\\
ac_{21} & ac_{22} & ac_{23} & ac_{24} \\\\
ac_{31} & ac_{32} & ac_{33} & ac_{34}
\\end{pmatrix}$

To work out the values for the answer matrix we multiply:-

Each value in row 1 in matrix $A$ by the corresponding value in column 1 in matrix $C$ – and work out the total. This provides the value for $ac_{11}$
Each value in row 1 in matrix $A$ by the corresponding value in column 2 in matrix $C$ – and work out the total. This provides the value for $ac_{12}$
Each value in row 1 in matrix $A$ by the corresponding value in column 3 in matrix $C$ – and work out the total. This provides the value for $ac_{13}$
Each value in row 1 in matrix $A$ by the corresponding value in column 4 in matrix $C$ – and work out the total. This provides the value for $ac_{14}$
Each value in row 2 in matrix $A$ by the corresponding value in column 1 in matrix $C$ – and work out the total. This provides the value for $ac_{21}$
and so on for each corresponding element in the answer matrix.

So:

$ac_{11} = a_{11}c_{11}+a_{12}c_{21} = \\var{maMa[0][0]}\\times\\var{maMc[0][0]} + \\var{maMa[0][1]}\\times\\var{maMc[1][0]} = \\var{maMac[0][0]}$

$ac_{12} = a_{11}c_{12}+a_{12}c_{22} = \\var{maMa[0][0]}\\times\\var{maMc[0][1]} + \\var{maMa[0][1]}\\times\\var{maMc[1][1]} = \\var{maMac[0][1]}$

$ac_{13} = a_{11}c_{13}+a_{12}c_{23} = \\var{maMa[0][0]}\\times\\var{maMc[0][2]} + \\var{maMa[0][1]}\\times\\var{maMc[1][2]} = \\var{maMac[0][2]}$

$ac_{14} = a_{11}c_{14}+a_{12}c_{24} = \\var{maMa[0][0]}\\times\\var{maMc[0][3]} + \\var{maMa[0][1]}\\times\\var{maMc[1][3]} = \\var{maMac[0][3]}$

$ac_{21} = a_{21}c_{11}+a_{22}c_{21} = \\var{maMa[1][0]}\\times\\var{maMc[0][0]} + \\var{maMa[1][1]}\\times\\var{maMc[1][0]} = \\var{maMac[1][0]}$

and so on...

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Scalar Multiplication

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Perform the following matrix scalar multiplications.

$\\simplify{{maS1}}\\times\\simplify{{rand1}} =$ [[0]]

$\\simplify{{maS2}}\\times\\simplify{{rand2}} =$ [[1]]

$\\simplify{{maS3}}\\times\\simplify{{rand3}} =$ [[2]]

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Matrix by Matrix Multiplication

Consider the following matrices

$A=\\simplify{{maMa}}$ $B=\\simplify{{maMb}}$ $C=\\simplify{{maMc}}$ $D=\\simplify{{maMd}}$ $E=\\simplify{{maMe}}$

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Which of the following combinations can be defined?

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What will the order of the matrix produced be for each of these multiplications?

AC: [[0]]x [[1]]

BA: [[2]]x [[3]]

CE: [[4]]x [[5]]

EC: [[6]]x [[7]]

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"unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "numcolumns(maMe*maMc)", "maxValue": "numcolumns(maMe*maMc)", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "matrix", "useCustomName": false, "customName": "", "marks": "4", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the product of AC.

AC = $\\simplify{{maMa}}\\times\\simplify{{maMc}} =$ ?

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Calculate the product of CE.

CE = $\\simplify{{maMc}}\\times\\simplify{{maMe}} =$ ?

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Calculate the product of EC.

EC = $\\simplify{{maMe}}\\times\\simplify{{maMc}} =$ ?

", "correctAnswer": "maMe*maMc", "correctAnswerFractions": false, "numRows": 1, "numColumns": 1, "allowResize": true, "tolerance": 0, "markPerCell": true, "allowFractions": false, "minColumns": 1, "maxColumns": 0, "minRows": 1, "maxRows": 0}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Matrix Determinants", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Mark Patterson", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/5064/"}], "tags": [], "metadata": {"description": "

Exercises on calculating the determinant of 2x2 and 3x3 matrices.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Matrix Determinants

Calculate the determinants of the following matrices.

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Calculate the determinant of $A = \\simplify{{maA}}$

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Calculate the determinant of $B = \\simplify{{maB}}$

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Calculate the determinant of $C = \\simplify{{maC}}$

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Calculate the determinant of $D = \\simplify{{maD}}$

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Exercises on inverting 2x2 and 3x3 matrices.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Matrix Inversion

Invert the following matrices. (Input your answers as fractions.)

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$A = \\simplify{{maA}}$

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$B = \\simplify{{maB}}$

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$C = \\simplify{{maC}}$

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$D = \\simplify{{maD}}$

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Exercises in solving simultaneous equations with 2 variables.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Simultaneous Equations

Solve the following simultaneous equations.

", "advice": "

To solve simultaneous equations, first rewrite the equations in matrix form of $Ab = C$, where the matrix $A$ will contain the coefficients of the equation, $b$ will contain the unknown values and $C$ will contain the constants.

We rearrange $Ab = C$ to get $A^{-1}C = b$.

So, we need to find the inverse of $A$ and multiple it with $C$. This will give us $b$, i.e. the values for $x$ and $y$.

For the first question:

$A = \\simplify{{maA}}$ $b = \\begin{pmatrix}x\\\\y\\end{pmatrix}$ $C = \\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix}$

To find the inverse of $A$, switch the elements on the leading diagonal, change the signs on the non-leading diagonal, and divide all elements by the determinant.

The determinant of $A$ is $|A| = (\\var{maA[0][0]}\\times\\var{maA[1][1]}) - (\\var{maA[0][1]}\\times\\var{maA[1][0]}) = \\var{detA}$

The inverse of $A$ is $A^{-1} = \\dfrac{1}{\\var{detA}}\\times\\begin{pmatrix}
\\var{maAA[0][0]} & \\var{maAA[0][1]} \\\\
\\var{maAA[1][0]} & \\var{maAA[1][1]}
\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}$

Rearranging for $A^{-1}C = b$:

$A^{-1}\\times C =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}} & \\frac{\\var{maAA[0][1]}}{\\var{detA}} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}} & \\frac{\\var{maAA[1][1]}}{\\var{detA}}
\\end{pmatrix}
\\times\\begin{pmatrix}\\var{a1}\\\\ \\var{a2}\\end{pmatrix} =
\\begin{pmatrix}
\\frac{\\var{maAA[0][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[0][1]}}{\\var{detA}}\\times\\var{a2} \\\\
\\frac{\\var{maAA[1][0]}}{\\var{detA}}\\times\\var{a1} + \\frac{\\var{maAA[1][1]}}{\\var{detA}}\\times\\var{a2}
\\end{pmatrix} =
\\begin{pmatrix}\\var{ax}\\\\ \\var{ay}\\end{pmatrix}$

Therefore $x = \\var{ax}$ and $y = \\var{ay}$.

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$\\simplify{{maA[0][0]}}x + \\simplify{{maA[0][1]}}y = \\simplify{{a1}}$

$\\simplify{{maA[1][0]}}x +\\simplify{{maA[1][1]}}y = \\simplify{{a2}}$

$x =$ [[0]]

$y =$ [[1]]

$\\simplify{{maB[0][0]}}x + \\simplify{{maB[0][1]}}y = \\simplify{{b1}}$

$\\simplify{{maB[1][0]}}x +\\simplify{{maB[1][1]}}y = \\simplify{{b2}}$

$x =$ [[0]]

$y =$ [[1]]

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