Inverse Matrix

Error

There was an error loading the page.

Contributors

History

○

Owen Jepps 7 years, 10 months ago

Created this as a copy of Inverse Matrix.

Name	Status	Author	Last Modified
Inverse Matrix	Ready to use	Christian Lawson-Perfect	01/06/2016 11:03
Habiba's copy of Inverse Matrix	Ready to use	Habiba Gora	01/06/2016 09:47
Ricardo's copy of Inverse Matrix	Ready to use	Ricardo Monge	01/06/2016 09:47
Hollie's copy of Inverse Matrix	draft	Hollie Tarr	01/12/2016 12:34
Hollie's copy of Inverse Matrix	draft	Hollie Tarr	22/02/2017 14:23
Inverse Matrix	draft	Owen Jepps	07/09/2020 01:38
Stephen's copy of (CLP's) Inverse Matrix	draft	Thomas Winyard	18/01/2022 21:07
Thomas's copy of Stephen's copy of (CLP's) Inverse Matrix	draft	Thomas Winyard	21/02/2019 19:16

There are 19 other versions that do you not have access to.

Question statement

Find the inverse of the following matrix:
$\simplify{matrix:A = {a}}$

Form the $3 \times 6$ augmented matrix $\mathrm{B}$ by placing $I_3$ to the right of $\mathrm{A}$ as below:

$\mathrm{B} = \left(\begin{array}{rrr|ccc} \var{a[0][0]} & \var{a[0][1]} & \var{a[0][2]} &\var{1}&\var{0}&\var{0}\\ \var{a[1][0]} & \var{a[1][1]} & \var{a[1][2]}&\var{0}&\var{1}&\var{0}\\ \var{a[2][0]} & \var{a[2][1]} & \var{a[2][2]}&\var{0}&\var{0}&\var{1}\\ \end{array}\right)$

In subsequent parts work with this matrix using row operations to introduce the identity matrix on the left hand side, with the inverse of $\mathrm{A}$ eventually appearing on the right hand side.

Input all numbers as fractions or integers and not as decimals.

Give any introductory information the student needs.

			Name	Type	Generated Value

matrix

Matrix of size 3×3

			Name	Type	Generated Value

step0	matrix	Matrix of size 3×6
zero_2_1	number	-3
zero_3_1	number	-9
one_2_2	number	1
step1	matrix	Matrix of size 3×6
zero_3_2	number	-18
one_3_3	number	-1
step2	matrix	Matrix of size 3×6
zero_1_3	number	6
zero_2_3	number	-3
step3	matrix	Matrix of size 3×6
zero_1_2	number	3
step4	matrix	Matrix of size 3×6
inverse	matrix	Matrix of size 3×3
one_3_3_message	string	Note that the third column in

			Name	Type	Generated Value

ra	integer	3
rb	integer	3
rc	integer	3
f	list	[ -1, -1, -1 ]
g	list	[ -1, 1, 1 ]

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
matrix([
  [f[0]*g[0],      f[0]*rb*g[1],         f[0]*(rb*ra-rb)*g[2]],
  [f[1]*ra*g[0],   f[1]*(ra*rb-1)*g[1],  f[1]*(ra^2*rb-ra-ra*rb)*g[2]],
  [f[2]*ra*rc*g[0],f[2]*rc*rb*g[1],      f[2]*g[2]]
])

JME function reference

Description

This matrix is designed so its determinant is $-1$ , meaning its inverse has integer components.

Additionally, when you introduce zeros in the second and third rows, you should end up with a $1$ in each of the leading columns, so you don't have to do a division.

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `matrix`

matrix([1,-3,-6],[3,-8,-15],[9,-9,-1])

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Matrix entry
  Add
Part b) Gap-fill
1. Gap Matrix entry
  Add
Part c) Gap-fill
1. Gap Matrix entry
  Add
Part d) Gap-fill
1. Gap Matrix entry
  Add

Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Introduce zeros in the first column below the first entry by adding suitable multiples of the first row to rows 2 and 3.

If necessary, multiply the second row by a suitable constant so that the second entry in the second row is $1$ .

$\left(\begin{array}{ccc|ccc} \var{step0[0][0]} & \var{step0[0][1]} & \var{step0[0][2]} & \var{step0[0][3]} & \var{step0[0][4]} & \var{step0[0][5]} \\ 0 & 1 & ? & ? & ? & ? \\ 0 & ? & ? & ? & ? & ? \end{array}\right) =$ Unnamed gap

Advice

a) Introduce zeros in the first column

The first entry in the second row is $\var{a[1][0]}$ , and we have a $1$ in the first entry of the first row, so multiply the first row by $\var{abs(zero_2_1)}$ and {if(zero_2_1<0,'subtract from','add to')} row 2 to get

$\simplify[rowvector,all]{{step0[1]}+{zero_2_1}*{step0[0]} = {step1[1]}}$

The first entry in the third row is $\var{a[2][0]}$ , so multiply the first row by $\var{abs(zero_3_1)}$ and {if(zero_3_1<0,'subtract from','add to')} row 3 to get

$\simplify[rowvector,all]{{step0[2]}+{zero_3_1}*{step0[0]} = {step1[2]}}$

The matrix at this stage is

$\var{step1}$

b) Introduce a zero in the second column of the third row

The second entry in the third row is $\var{step1[2][1]}$ , and we have a $1$ in the second entry of the second row, so multiply the second row by $\var{abs(zero_3_2)}$ and {if(zero_3_2<0,'subtract from','add to')} row 3 to get

$\simplify[rowvector,all]{{step1[2]}+{zero_3_2}*{step1[1]} = {step1[2]+zero_3_2*step1[1]}}$

{if(one_3_3<>1,one_3_3_message,'')}

The matrix at this stage is

$\var{step2}$

c) Introduce zeros in the third column

The third entry in the first row is $\var{step2[0][2]}$ , and we have a $1$ in the third entry of the third row, so multiply the third row by $\var{abs(zero_1_3)}$ and {if(zero_1_3<0,'subtract from','add to')} row 1 to get

$\simplify[rowvector,all]{{step2[0]}+{zero_1_3}*{step2[2]} = {step2[0]+zero_1_3*step2[2]}}$

The third entry in the second row is $\var{step2[1][2]}$ , so multiply the third row by $\var{abs(zero_2_3)}$ and {if(zero_2_3<0,'subtract from','add to')} row 2 to get

$\simplify[rowvector,all]{{step2[1]}+{zero_2_3}*{step2[2]} = {step2[1]+zero_2_3*step2[2]}}$

The matrix at this stage is

$\var{step3}$

d) Introduce a zero in the second column of the first row

The second entry in the first row is $\var{step3[0][1]}$ , and we have a $1$ in the second entry of the second row, so multiply the second row by $\var{abs(zero_1_2)}$ and {if(zero_1_2<0,'subtract from','add to')} row 1 to get

$\simplify[rowvector,all]{{step3[0]}+{zero_1_2}*{step3[1]} = {step3[0]+zero_1_2*step3[1]}}$

The matrix at this stage is

$\var{step4}$

Note that we have a $3 \times 3$ identity matrix in the left three columns, and $\mathrm{A}^{-1}$ is written in the three columns on the right. That is,

$\mathrm{A}^{-1} = \var{inverse}$

Give a worked solution to the whole question.

Use this tab to check that this question works as expected.

There was an error which means the tests can't run:

Part	Test	Passed?
Gap-fill
Gap-fill		Hasn't run yet
Matrix entry
Matrix entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Matrix entry
Matrix entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Matrix entry
Matrix entry		Hasn't run yet
Gap-fill
Gap-fill		Hasn't run yet
Matrix entry
Matrix entry		Hasn't run yet

This question is used in the following exams:

Matrices (Second Week) by Owen Jepps in 1010ENG/1201SCG Matrices.
Isaac's copy of Review Questions by Isaac Mulolani in Isaac's workspace.
Maths 1A - Homework Week 3 (Matrices) by Maria Aneiros in MY QUESTIONS.
Review Questions by Isaac Mulolani in Isaac's workspace.
Week 7 - Matrices 2 by Peter Johnston in 1010ENG/1201SCG Matrices.

Loading...

Error

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Owen Jepps 7 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: `matrix`

Parts

Gap-fill

a) Introduce zeros in the first column

b) Introduce a zero in the second column of the third row

c) Introduce zeros in the third column

d) Introduce a zero in the second column of the first row

Loading...

Error

Inverse Matrix

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Owen Jepps 7 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: matrix

Parts

Gap-fill

a) Introduce zeros in the first column

b) Introduce a zero in the second column of the third row

c) Introduce zeros in the third column

d) Introduce a zero in the second column of the first row

Generated value: `matrix`