Mark's copy of Inverse of a 3x3 matrix

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Mark Hodds 7 years, 7 months ago

Created this as a copy of Inverse of a 3x3 matrix.

Name	Status	Author	Last Modified
Inverse of a 3x3 matrix	Ready to use	Frank Doheny	14/03/2018 21:05
Mark's copy of Inverse of a 3x3 matrix	draft	Mark Hodds	07/09/2017 14:58
Mark's copy of Mark's copy of Inverse of a 3x3 matrix	draft	Mark Hodds	07/09/2017 15:00
Determinant of a 3x3 matrix	Ready to use	Frank Doheny	14/03/2018 21:10
Determinant of a 3x3 matrix	draft	Xiaodan Leng	11/07/2019 00:31
3x3 Determinant Problems	draft	Tamsin Smith	17/09/2024 13:14

There are 5 other versions that do you not have access to.

Question statement

Given the matrix:

$\mathbf{A}=\begin{pmatrix} \var{a11}&\var{a12}&\var{a13}\\ \var{a21}&\var{a22}&\var{a23}\\\var{a31}&\var{a32}&\var{a33} \end{pmatrix}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a11	integer	2
a12	integer	4
a13	integer	2
k	integer	5
a21	integer	10
a22	integer	21
a23	integer	4
k1	integer	6
a31	integer	12
a32	integer	30
a33	integer	-23
b12	rational	76
b13	rational	-13
b21	rational	139
b22	rational	-35
b23	rational	6
b31	rational	24
b32	rational	-6
b33	rational	1
b11	rational	-603/2
r1	integer	1
s1	integer	5
t1	integer	8
r	integer	38
s	integer	147
t	integer	-22

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Generated value: `integer`

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Parts

Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Matrix entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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5. Gap Gap 4. Number entry
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Gap-fill

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Calculate the determinant of the matrix.

$|A|=$ Gap 0.

Calculate the inverse matrix and input the entries correct to two decimal places.

$A^{-1}=$ Gap 1.

Hence solve the following system of equations.

$\begin{pmatrix} \var{a11}&\var{a12}&\var{a13}\\ \var{a21}&\var{a22}&\var{23}\\ \var{a31}&\var{a32}&\var{a33}\end{pmatrix}\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix} \var{r}\\\var{s}\\\var{t}\end{pmatrix}$

$x=$ Gap 2.

$y=$ Gap 3.

$z=$ Gap 4.

Advice

Given the matrix:

$\mathbf{A}=\begin{pmatrix} {a11}&{a12}&{a13}\\ {a21}&{a22}&{a23}\\{a31}&{a32}&{a33} \end{pmatrix}$

The determinant of a 3x3 matrix is determined by the formula:

$|\mathbf{A}|={a11}*\begin{vmatrix}{a22}&{a23}\\{a32}&{a33}\end{vmatrix}-{a12}*\begin{vmatrix}{a21}&{a23}\\{a31}&{a33}\end{vmatrix}+{a13}*\begin{vmatrix}{a21}&{a22}\\{a31}&{a32}\end{vmatrix}$

So in this example:

$\mathbf{A}=\begin{pmatrix} \var{a11}&\var{a12}&\var{a13}\\ \var{a21}&\var{a22}&\var{a23}\\\var{a31}&\var{a32}&\var{a33} \end{pmatrix}$

$|\mathbf{A}|=\var{a11}*\begin{vmatrix}\var{a22}&\var{a23}\\\var{a32}&\var{a33}\end{vmatrix}-\var{a12}*\begin{vmatrix}\var{a21}&\var{a23}\\\var{a31}&\var{a33}\end{vmatrix}+\var{a13}*\begin{vmatrix}\var{a21}&\var{a22}\\\var{a31}&\var{a32}\end{vmatrix}$

$|\mathbf{A}|=\var{a11}*(\var{a22}*\var{a33}-\var{a32}*\var{a23})-\var{a12}*(\var{a21}*\var{a33}-\var{a31}*\var{a23})+\var{a13}*(\var{a21}*\var{a32}-\var{a31}*\var{a22})$

$|\mathbf{A}|=(\simplify{{a11}*({a22}*{a33}-{a32}*{a23})})-(\simplify{{a12}*({a21}*{a33}-{a31}*{a23})})+(\simplify{{a13}*({a21}*{a32}-{a31}*{a22})})$

$|\mathbf{A}|=\var{a11}$

The adjoint of the matrix is given by:

$\mathbf{A}^*=\begin{pmatrix}\begin{vmatrix}\var{a22}&\var{a23}\\\var{a32}&\var{a33}\end{vmatrix}&-\begin{vmatrix}\var{a21}&\var{a23}\\\var{a31}&\var{a33}\end{vmatrix}&\begin{vmatrix}\var{a21}&\var{a22}\\\var{a31}&\var{a32}\end{vmatrix}\\-\begin{vmatrix}\var{a12}&\var{a13}\\\var{a32}&\var{a33}\end{vmatrix}&\begin{vmatrix}\var{a11}&\var{a13}\\\var{a31}&\var{a33}\end{vmatrix}&-\begin{vmatrix}\var{a11}&\var{a12}\\\var{a31}&\var{a32}\end{vmatrix}\\\begin{vmatrix}\var{a12}&\var{a13}\\\var{a22}&\var{a23}\end{vmatrix}&-\begin{vmatrix}\var{a11}&\var{a13}\\\var{a21}&\var{a23}\end{vmatrix}&\begin{vmatrix}\var{a11}&\var{a12}\\\var{a21}&\var{a11}\end{vmatrix}\end{pmatrix}^t$

$\mathbf{A}^*=\begin{pmatrix}\simplify{({a22}*{a33}-{a32}*{a23})}&\simplify{({a23}*{a31}-{a21}*{a33})}&\simplify{({a21}*{a32}-{a22}*{a31})}\\\simplify{(-{a12}*{a33}+{a32}*{a13})}&\simplify{({a11}*{a33}-{a13}*{a31})}&\simplify{({a12}*{a31}-{a32}*{a11})}\\\simplify{({a12}*{a23}-{a22}*{a13})}&\simplify{({a21}*{a13}-{a11}*{a23})}&\simplify{({a22}*{a11}-{a12}*{a21})}\end{pmatrix}^t$

$\mathbf{A}^*=\begin{pmatrix}\simplify{({a22}*{a33}-{a32}*{a23})}&\simplify{(-{a12}*{a33}+{a32}*{a13})}&\simplify{({a12}*{a23}-{a22}*{a13})}\\\simplify{({a23}*{a31}-{a21}*{a33})}&\simplify{({a11}*{a33}-{a13}*{a31})}&\simplify{({a21}*{a13}-{a11}*{a23})}\\\simplify{({a21}*{a32}-{a22}*{a31})}&\simplify{({a12}*{a31}-{a32}*{a11})}&\simplify{({a22}*{a11}-{a12}*{a21})}\end{pmatrix}$

The inverse matrix is defined by $A^{-1}=\frac{1}{|A|}A^*$

$A^{-1}=\frac{1}{\var{a11}}\begin{pmatrix}\simplify{({a22}*{a33}-{a32}*{a23})}&\simplify{(-{a12}*{a33}+{a32}*{a13})}&\simplify{({a12}*{a23}-{a22}*{a13})}\\\simplify{({a23}*{a31}-{a21}*{a33})}&\simplify{({a11}*{a33}-{a13}*{a31})}&\simplify{({a21}*{a13}-{a11}*{a23})}\\\simplify{({a21}*{a32}-{a22}*{a31})}&\simplify{({a12}*{a31}-{a32}*{a11})}&\simplify{({a22}*{a11}-{a12}*{a21})}\end{pmatrix}$

Using the inverse matrix method gives:

$\begin{pmatrix} x\\y\\z\end{pmatrix}=\frac{1}{\var{a11}}\begin{pmatrix}\simplify{({a22}*{a33}-{a32}*{a23})}&\simplify{(-{a12}*{a33}+{a32}*{a13})}&\simplify{({a12}*{a23}-{a22}*{a13})}\\\simplify{({a23}*{a31}-{a21}*{a33})}&\simplify{({a11}*{a33}-{a13}*{a31})}&\simplify{({a21}*{a13}-{a11}*{a23})}\\\simplify{({a21}*{a32}-{a22}*{a31})}&\simplify{({a12}*{a31}-{a32}*{a11})}&\simplify{({a22}*{a11}-{a12}*{a21})}\end{pmatrix}\begin{pmatrix} \var{r}\\\var{s}\\\var{t}\end{pmatrix}$

$\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix}\var{b11}&\var{b12}&\var{b13}\\\var{b21}&\var{b22}&\var{b23}\\\var{b31}&\var{b32}&\var{b33}\end{pmatrix}\begin{pmatrix} \var{r}\\\var{s}\\\var{t}\end{pmatrix}$

$\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix}(\var{b11})*\var{r}+(\var{b12})*\var{s}+(\var{b13})*\var{t}\\(\var{b21})*\var{r}+(\var{b22})*\var{s}+(\var{b23})*\var{t}\\(\var{b31})*\var{r}+(\var{b32})*\var{s}+(\var{b33})*\var{t}\end{pmatrix}$

$\begin{pmatrix} x\\y\\z\end{pmatrix}=\begin{pmatrix} \var{r1}\\\var{s1}\\\var{t1}\end{pmatrix}$

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This question is used in the following exam:

101MP 2019 - Matrices Week 4 by Mark Hodds in 101MP 2018.

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Mark Hodds 7 years, 7 months ago

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Mark's copy of Inverse of a 3x3 matrix

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Mark Hodds 7 years, 7 months ago

Error in variable testing condition

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Generated value: integer

Parts

Gap-fill

Generated value: `integer`