// Numbas version: exam_results_page_options {"name": "Matrices: Determinants (Instructional)", "metadata": {"description": "

2x2 by ad-bc, 3x3 by Laplace expansion, properties of determinants

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Determinant of 2x2 and notation

\n

Students are asked to form the calculation before giving answer.

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

The determinant of the matrix:

\n

$\\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| $

\n

and 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 are asked to calculate the determinants of some  $2 \\times 2$  matrices:

\n

 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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Calculate the determinants of the following matrices:

\n

 

\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

\n

Input answer only.

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

The determinant of the matrix:

\n

$\\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| $

\n

and 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 are asked to calculate the determinants of some  $2 \\times 2$  matrices:

\n

 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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Calculate the determinants of the following matrices:

\n

 

\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": "

The determinant of a square matrix can be obtained by multiplying each element of a chosen row by its corresponding cofactor and then adding these products together.

\n

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}\\]

\n

This technique is known as Laplace Expansion.

", "advice": "

We are asked to use Laplace expansion along the $1^{st}$, top row to calculate the determinant:

\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| \\]

\n

In this case, the elements in the first row are:

\n

$a_{11}=\\var{a11}$          $a_{12}=\\var{a12}$           $a_{13}=\\var{a13}$

\n

\n

Remember, 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\n\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

\n

Now 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

 

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Use Laplace Expansion along the $1^{st}$, top row to calculate the determinant:

\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| \\]

\n

In this case, the elements in the first row are:

\n

$a_{11}=$ [[0]]          $a_{12}=$ [[1]]           $a_{13}=$ [[2]]

\n

\n

And so the minors with their co-factors are:

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\LARGE{A_{11}=+}$\n

$\\Huge{|}$

\n
[[3]][[4]]\n

$\\Huge{|}$

\n
$=$[[7]]
[[5]][[6]]
\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\LARGE{A_{12}=-}$\n

$\\Huge{|}$

\n
[[8]][[9]]\n

$\\Huge{|}$

\n
$=$[[12]]
[[10]][[11]]
\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\LARGE{A_{13}=+}$\n

$\\Huge{|}$

\n
[[13]][[14]]\n

$\\Huge{|}$

\n
$=$[[17]]
[[15]][[16]]
\n

\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]]

\n

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Useful properties of determinants that allow simplification.

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", "licence": "Creative Commons Attribution-NonCommercial-NoDerivs 4.0 International"}, "statement": "

Determinants exhibit some properties that can usefully be thought of as \"Rules\":

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Rule 1:  If two rows (or two columns) of a determinant are interchanged then the value of the determinant is multiplied by ($−1$).

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Rule 2:  The determinant of a matrix $A$ and the determinant of its transpose $A^T$ are equal.

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Rule 3:  If two rows (or two columns) of a matrix $A$ are equal then it has zero determinant.

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Rule 4:  If the elements of one row (or one column) of a determinant are multiplied by $k$, then the determinant is also multiplied by $k$.

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Rule 5:  If we add (or subtract) a multiple of one row (or column) to another, the value of the determinant is unchanged.

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Rule 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": "

We are asked to give the determinants for various matrices without explicitly calculating them. We can do this by applying the \"rules\" given:

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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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Row $1$ and Row $3$ are equal so the determinant is $0$ (by Rule 3).

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 $\\large \\left| \\begin{array}{ccc} \\var{a11} & 0 & 0 & 0\\\\  \\var{a12} &  \\var{a11} & 0 & 0 \\\\ \\var{a22}& \\var{a21}& 1 & 0 \\\\ \\var{a12} & 0 & \\var{a21} & \\var{a22} \\end{array} \\right| =\\simplify{{a11}{a11}{a22}}$

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Calculating 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).

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If          $\\large \\left| \\begin{array}{ccc} \\var{a11} & \\var{a12}  \\\\ \\var{a21}& \\var{a22}  \\end{array} \\right| =\\var{detA1}$          then          $\\large \\left| \\begin{array}{ccc} \\var{a21} & \\var{a22}  \\\\ \\var{a11}& \\var{a12}  \\end{array} \\right| =\\var{detA2}$

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In 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).

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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     $\\left|\\begin{array}{ccc}\\var{B[0][0]} & \\var{B[1][0]}& \\var{B[2][0]}\\\\ \\var{B[0][1]} & \\var{B[1][1]}& \\var{B[2][1]}\\\\ \\var{B[0][2]} & \\var{B[1][2]}& \\var{B[2][2]}\\end{array}\\right|=\\var{detB}$

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You should, hopefully, be able to see that the second matrix is the transpose of the first. Hence their determinants are equal (by Rule 2).

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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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