// Numbas version: finer_feedback_settings {"name": "Matrices: Determinants (Instructional)", "metadata": {"description": "
2x2 by ad-bc, 3x3 by Laplace expansion, properties of determinants
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\nStudents are asked to form the calculation before giving answer.
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "$\\large A= \\left( \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right) $ is denoted by $\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| $
\nand is defined to be the number $ad-bc$. That is:
\n$\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| =ad-bc$
\n\n
We can use the notation $det(A)$ or $|A|$ or $\\Delta$ to denote the determinant of $A$.
", "advice": "We simply have to carry out this calculation for each one,
\n$\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| =ad-bc$
\n$A=\\var{A}$
\n$|A|=\\left|\\begin{array}{ccc}\\var{A[0][0]} & \\var{A[0][1]} \\\\\\var{A[1][0]} & \\var{A[1][1]} \\end{array}\\right| =\\var{Apart1}\\large -\\var{Apart2}=\\var{detA}$
\n\n
$B=\\var{B}$
\n$|B|=\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} \\\\\\var{B[1][0]} & \\var{B[1][1]} \\end{array}\\right| =\\var{Bpart1}\\large -\\var{Bpart2}=\\var{detB}$
\n\n
$C=\\var{C}$
\n$|C|= \\left|\\begin{array}{ccc}\\var{C[0][0]} & \\var{C[0][1]} \\\\\\var{C[1][0]} & \\var{C[1][1]} \\end{array}\\right| =\\var{Cpart1}\\large -\\var{Cpart2}=\\var{detC}$
\n\n
$D=\\var{D}$
\n$|D|= \\left|\\begin{array}{ccc}\\var{D[0][0]} & \\var{D[0][1]} \\\\\\var{D[1][0]} & \\var{D[1][1]} \\end{array}\\right| =\\var{Dpart1}\\large -\\var{Dpart2}=\\var{detD}$
\n\n
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\n
$A=\\var{A}$
\n$|A|=\\left|\\begin{array}{ccc}\\var{A[0][0]} & \\var{A[0][1]} \\\\\\var{A[1][0]} & \\var{A[1][1]} \\end{array}\\right| =$ [[0]]$\\large -$[[1]]$=$ [[2]]
\n\n
$B=\\var{B}$
\n$|B|=\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} \\\\\\var{B[1][0]} & \\var{B[1][1]} \\end{array}\\right| =$ [[3]]$\\large -$[[4]]$=$ [[5]]
\n\n
$C=\\var{C}$
\n$|C|= \\left|\\begin{array}{ccc}\\var{C[0][0]} & \\var{C[0][1]} \\\\\\var{C[1][0]} & \\var{C[1][1]} \\end{array}\\right| =$ [[6]]$\\large -$[[7]]$=$ [[8]]
\n\n
$D=\\var{D}$
\n$|D|= \\left|\\begin{array}{ccc}\\var{D[0][0]} & \\var{D[0][1]} \\\\\\var{D[1][0]} & \\var{D[1][1]} \\end{array}\\right| =$ [[9]]$\\large -$[[10]]$=$ [[11]]
\n\n
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false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{Bpart2}", "maxValue": "{Bpart2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{detB}", "maxValue": "{detB}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", 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Determinant of 2x2 and notation
\nInput answer only.
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "$\\large A= \\left( \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right) $ is denoted by $\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| $
\nand is defined to be the number $ad-bc$. That is:
\n$\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| =ad-bc$
\n\n
We can use the notation $det(A)$ or $|A|$ or $\\Delta$ to denote the determinant of $A$.
", "advice": "We simply have to carry out this calculation for each one,
\n$\\large \\left| \\begin{array}{ccc} a & b \\\\ c & d \\end{array} \\right| =ad-bc$
\n$A=\\var{A}$
\n$|A|=\\left|\\begin{array}{ccc}\\var{A[0][0]} & \\var{A[0][1]} \\\\\\var{A[1][0]} & \\var{A[1][1]} \\end{array}\\right| =\\var{Apart1}\\large -\\var{Apart2}=\\var{detA}$
\n\n
$B=\\var{B}$
\n$|B|=\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} \\\\\\var{B[1][0]} & \\var{B[1][1]} \\end{array}\\right| =\\var{Bpart1}\\large -\\var{Bpart2}=\\var{detB}$
\n\n
$C=\\var{C}$
\n$|C|= \\left|\\begin{array}{ccc}\\var{C[0][0]} & \\var{C[0][1]} \\\\\\var{C[1][0]} & \\var{C[1][1]} \\end{array}\\right| =\\var{Cpart1}\\large -\\var{Cpart2}=\\var{detC}$
\n\n
$D=\\var{D}$
\n$|D|= \\left|\\begin{array}{ccc}\\var{D[0][0]} & \\var{D[0][1]} \\\\\\var{D[1][0]} & \\var{D[1][1]} \\end{array}\\right| =\\var{Dpart1}\\large -\\var{Dpart2}=\\var{detD}$
\n\n
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\n
$A=\\var{A}$
\n$|A|=\\left|\\begin{array}{ccc}\\var{A[0][0]} & \\var{A[0][1]} \\\\\\var{A[1][0]} & \\var{A[1][1]} \\end{array}\\right| =$ [[0]]
\n\n
$B=\\var{B}$
\n$|B|=\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} \\\\\\var{B[1][0]} & \\var{B[1][1]} \\end{array}\\right| =$ [[1]]
\n\n
$C=\\var{C}$
\n$|C|=\\left|\\begin{array}{ccc}\\var{C[0][0]} & \\var{C[0][1]} \\\\\\var{C[1][0]} & \\var{C[1][1]} \\end{array}\\right|=$ [[2]]
\n\n
$D=\\var{D}$
\n$|D|= \\left|\\begin{array}{ccc}\\var{D[0][0]} & \\var{D[0][1]} \\\\\\var{D[1][0]} & \\var{D[1][1]} \\end{array}\\right| =$ [[3]]
\n\n
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Determinant of n x m matrix by Laplace Expansion across top row.
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "For example, chosing the first row of a $3 \\times 3$ matrix:
\n\\[ \\Large \\Delta=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}\\]
\nThis technique is known as Laplace Expansion.
", "advice": "\\[ \\Large{\\Delta}=\\left|\\begin{array}{ccc}\\var{a11} & \\var{a12} & \\var{a13} \\\\ \\var{a21} & \\var{a22} & \\var{a23} \\\\ \\var{a31} & \\var{a32} & \\var{a33} \\end{array}\\right| \\]
\nIn this case, the elements in the first row are:
\n$a_{11}=\\var{a11}$ $a_{12}=\\var{a12}$ $a_{13}=\\var{a13}$
\n\nRemember, the minor for each element is formed by \"striking out\" the row and column that the element appears in, leaving a smaller matrix. The co-factor for each one is positive or negative following the pattern:
\n$+$ | \n$-$ | \n$+$ | \n
$-$ | \n$+$ | \n$-$ | \n
$+$ | \n$-$ | \n$+$ | \n
And so the minors with their co-factors are:
\n$\\LARGE{A_{11}=+}\\left| \\begin{array}{ccc} \\var{CF11[0][0]} & \\var{CF11[0][1]} \\\\ \\var{CF11[1][0]} & \\var{CF11[1][1]}\\end{array}\\right|=\\var{detCF11}$
\n\n
$\\LARGE{A_{12}=+}\\left| \\begin{array}{ccc} \\var{CF12[0][0]} & \\var{CF12[0][1]} \\\\ \\var{CF12[1][0]} & \\var{CF12[1][1]}\\end{array}\\right|=-\\var{detCF12}$
\n\n
$\\LARGE{A_{13}=+}\\left| \\begin{array}{ccc} \\var{CF13[0][0]} & \\var{CF13[0][1]} \\\\ \\var{CF13[1][0]} & \\var{CF13[1][1]}\\end{array}\\right|=\\var{detCF13}$
\n\n\n\nNow multiply each element by its corresponding cofactor/minor and add these product together:
\n\n
$ \\Large{\\Delta} =\\Large{(}\\var{a11}\\times\\var{detCF11}\\Large{)+(}\\var{a12}\\times-\\var{detCF12}\\Large{)+(}\\var{a13}\\times\\var{detCF13}\\Large{)}=\\var{detA}$
\n\n
\\[ \\Large{\\Delta}=\\left|\\begin{array}{ccc}\\var{a11} & \\var{a12} & \\var{a13} \\\\ \\var{a21} & \\var{a22} & \\var{a23} \\\\ \\var{a31} & \\var{a32} & \\var{a33} \\end{array}\\right| \\]
\nIn this case, the elements in the first row are:
\n$a_{11}=$ [[0]] $a_{12}=$ [[1]] $a_{13}=$ [[2]]
\n\nAnd so the minors with their co-factors are:
\n\n$\\LARGE{A_{11}=+}$ | \n\n$\\Huge{|}$\n | \n[[3]] | \n[[4]] | \n\n$\\Huge{|}$\n | \n$=$ | \n[[7]] | \n
[[5]] | \n[[6]] | \n
$\\LARGE{A_{12}=-}$ | \n\n$\\Huge{|}$\n | \n[[8]] | \n[[9]] | \n\n$\\Huge{|}$\n | \n$=$ | \n[[12]] | \n
[[10]] | \n[[11]] | \n
$\\LARGE{A_{13}=+}$ | \n\n$\\Huge{|}$\n | \n[[13]] | \n[[14]] | \n\n$\\Huge{|}$\n | \n$=$ | \n[[17]] | \n
[[15]] | \n[[16]] | \n
Now multiply each element by its corresponding cofactor/minor and add these product together:
\n\n
$ \\Large{\\Delta} =\\Large{(}$[[18]]$\\times$[[19]]$\\Large{)+(}$[[20]]$\\times$[[21]]$\\Large{)+(}$[[22]]$\\times$[[23]]$\\Large{)}=$[[24]]
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\nRule 1: If two rows (or two columns) of a determinant are interchanged then the value of the determinant is multiplied by ($−1$).
\nRule 2: The determinant of a matrix $A$ and the determinant of its transpose $A^T$ are equal.
\nRule 3: If two rows (or two columns) of a matrix $A$ are equal then it has zero determinant.
\nRule 4: If the elements of one row (or one column) of a determinant are multiplied by $k$, then the determinant is also multiplied by $k$.
\nRule 5: If we add (or subtract) a multiple of one row (or column) to another, the value of the determinant is unchanged.
\nRule 6: The determinant of a lower triangular matrix, an upper triangular matrix or a diagonal matrix is the product of the elements on the leading diagonal.
", "advice": "\n
$\\large \\left| \\begin{array}{ccc} \\var{a22} & \\var{a11} & \\var{a21}\\\\ \\var{a11} & -7 & \\var{a11} \\\\\\var{a22}& \\var{a11} & \\var{a21} \\end{array} \\right| =0$
\nRow $1$ and Row $3$ are equal so the determinant is $0$ (by Rule 3).
\n\n
\n
$\\large \\left| \\begin{array}{ccc} \\var{a11} & 0 & 0 & 0\\\\ \\var{a12} & \\var{a11} & 0 & 0 \\\\ \\var{a22}& \\var{a21}& 1 & 0 \\\\ \\var{a12} & 0 & \\var{a21} & \\var{a22} \\end{array} \\right| =\\simplify{{a11}{a11}{a22}}$
\nCalculating the determinant of matrices larger than $3 \\times 3$ can be a laborious process. However, this matrix is triangular so its determinant is merely the product of the values in the leading diagonal (by Rule 6).
\n\n
\n
If $\\large \\left| \\begin{array}{ccc} \\var{a11} & \\var{a12} \\\\ \\var{a21}& \\var{a22} \\end{array} \\right| =\\var{detA1}$ then $\\large \\left| \\begin{array}{ccc} \\var{a21} & \\var{a22} \\\\ \\var{a11}& \\var{a12} \\end{array} \\right| =\\var{detA2}$
\nIn this case, the second matrix has the same elements as the first but Row $1$ and Row $2$ have been interchanged. So the determinant of the second will be the same as the determinant of the first but $\\times -1$ (by Rule 1).
\n\n
\n
If $\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} & \\var{B[0][2]}\\\\ \\var{B[1][0]} & \\var{B[1][1]} & \\var{B[1][2]}\\\\ \\var{B[2][0]} & \\var{B[2][1]}& \\var{B[2][2]}\\end{array}\\right|=\\var{detB}$ then $\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[1][0]}& \\var{B[2][0]}\\\\ \\var{B[0][1]} & \\var{B[1][1]}& \\var{B[2][1]}\\\\ \\var{B[0][2]} & \\var{B[1][2]}& \\var{B[2][2]}\\end{array}\\right|=\\var{detB}$
\nYou should, hopefully, be able to see that the second matrix is the transpose of the first. Hence their determinants are equal (by Rule 2).
\n\n
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By using the properties of determinants (above), that is without calculating any determinants, answer the following:
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$\\large \\left| \\begin{array}{ccc} \\var{a22} & \\var{a11} & \\var{a21}\\\\ \\var{a11} & -7 & \\var{a11} \\\\\\var{a22}& \\var{a11} & \\var{a21} \\end{array} \\right| =$ [[0]]
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$\\large \\left| \\begin{array}{ccc} \\var{a11} & 0 & 0 & 0\\\\ \\var{a12} & \\var{a11} & 0 & 0 \\\\ \\var{a22}& \\var{a21}& 1 & 0 \\\\ \\var{a12} & 0 & \\var{a21} & \\var{a22} \\end{array} \\right| =$ [[1]]
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If $\\large \\left| \\begin{array}{ccc} \\var{a11} & \\var{a12} \\\\ \\var{a21}& \\var{a22} \\end{array} \\right| =\\var{detA1}$ then $\\large \\left| \\begin{array}{ccc} \\var{a21} & \\var{a22} \\\\ \\var{a11}& \\var{a12} \\end{array} \\right| =$ [[2]]
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If $\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[0][1]} & \\var{B[0][2]}\\\\ \\var{B[1][0]} & \\var{B[1][1]} & \\var{B[1][2]}\\\\ \\var{B[2][0]} & \\var{B[2][1]}& \\var{B[2][2]}\\end{array}\\right|=\\var{detB}$ then $\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[1][0]}& \\var{B[2][0]}\\\\ \\var{B[0][1]} & \\var{B[1][1]}& \\var{B[2][1]}\\\\ \\var{B[0][2]} & \\var{B[1][2]}& \\var{B[2][2]}\\end{array}\\right|=$ [[3]]
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