Simon's copy of Combine linearly dependent vectors to get zero

Error

There was an error loading the page.

Contributors

History

○

Simon Thomas 5 years, 10 months ago

Created this as a copy of Combine linearly dependent vectors to get zero.

Name	Status	Author	Last Modified
Combine linearly dependent vectors to get zero	Ready to use	Newcastle University Mathematics and Statistics	13/10/2024 20:14
Simon's copy of Combine linearly dependent vectors to get zero	draft	Simon Thomas	12/06/2019 14:17

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

Question statement

You are given the following four vectors in $\mathbb{R}^4$ :

$\begin{align} \textbf{v}_1&=\var{v1}\\ \textbf{v}_2&=\var{v2}\\ \textbf{v}_3&=\var{v3}\\ \textbf{v}_4&=\var{v4}\end{align}$

Show that the vectors $\textbf{v}_1,\;\textbf{v}_2,\;\textbf{v}_3,\;\textbf{v}_4$ are linearly dependent by finding real numbers $a,\;b,\;c$ and $d$ such that

$a\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3+d\textbf{v}_4=\textbf{0},\;\;\; a+b+c+d=1$ where

$\textbf{0}=\var{rowvector(0,0,0,0)}$ is the zero vector.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	3
x	matrix	matrix([2,-5,6,-1])
c	integer	-3
b	integer	2
ga	integer	2
al	integer	2
y	matrix	matrix([5,-2,0,5])
v1	matrix	matrix([-14,-2,16,-20])
v2	matrix	matrix([2,-5,6,-1])
v3	matrix	matrix([5,-2,0,5])
v4	matrix	matrix([1,0,2,1])
t	integer	1
w	matrix	matrix([-14,-2,16,-20])
c2	rational	-2
c3	rational	4
del	integer	0
c1	rational	1
z	matrix	matrix([1,0,2,1])
c4	rational	-2

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(-3..3 except 0)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

→ Used by:

Parts

Gap-fill
1. Gap Gap 0. Number entry
  Add
2. Gap Gap 1. Number entry
  Add
3. Gap Gap 2. Number entry
  Add
4. Gap Gap 3. Number entry
  Add

Gap-fill

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

$a=$ Gap 0.

$b=$ Gap 1.

$c=$ Gap 2.

$d=$ Gap 3.

Input all values as exact decimals.

Advice

On putting $a=1-b-c-d$ we have

$(1-b-c-d)\textbf{v}_1+b\textbf{v}_2+c\textbf{v}_3+d\textbf{v}_4=0$

Hence on combining the vectors and equating the four components each to $0$ we have the four equations:

$\begin{align} \simplify[std]{{v1[0][0]} * (1 -b -c -d) + {v2[0][0]} * b + {v3[0][0]} * c + {v4[0][0]} * d }&= 0\\ \simplify[std]{{v1[0][1]} * (1 -b -c -d) + {v2[0][1]} * b + {v3[0][1]} * c + {v4[0][1]} * d }&= 0\\ \simplify[std]{{v1[0][2]} * (1 -b -c -d) + {v2[0][2]} * b + {v3[0][2]} * c + {v4[0][2]} * d }&= 0\\ \simplify[std]{{v1[0][3]} * (1 -b -c -d) + {v2[0][3]} * b + {v3[0][3]} * c + {v4[0][3]} * d }&= 0\end{align}$

On rearranging these equations in the unknowns $b,\;c,\;d$ we get:

$\begin{align} \simplify[std]{{v2[0][0] -v1[0][0]} * b + {v3[0][0] -v1[0][0]} * c + {v4[0][0] -v1[0][0]} * d} &= \var{-v1[0][0]}\\ \simplify[std]{{v2[0][1] -v1[0][1]} * b + {v3[0][1] -v1[0][1]} * c + {v4[0][1] -v1[0][1]} * d }&= \var{-v1[0][1]}\\ \simplify[std]{{v2[0][2] -v1[0][2]} * b + {v3[0][2] -v1[0][2]} * c + {v4[0][2] -v1[0][2]} * d }&= \var{-v1[0][2]}\\ \simplify[std]{{v2[0][3] -v1[0][3]} * b + {v3[0][3] -v1[0][3]} * c + {v4[0][3] -v1[0][3]} * d}&= \var{-v1[0][3]}\end{align}$

We have 4 equations in 3 unknowns. However, we can solve the first 3 equations simultaneously. One method is by writing the equations in matrix form and inverting the $3 \times 3$ matrix obtained (an alternative method is Gaussian elimination):

$\begin{pmatrix} \var{v2[0][0] -v1[0][0]}& \var{v3[0][0] -v1[0][0]} & \var{v4[0][0] -v1[0][0]}\\ \var{v2[0][1] -v1[0][1]}&\var{v3[0][1] -v1[0][1]} &\var{v4[0][1] -v1[0][1]} \\ \var{v2[0][2] -v1[0][2]}& \var{v3[0][2] -v1[0][2]}&\var{v4[0][2] -v1[0][2]} \end{pmatrix} \begin{pmatrix} b \\ c \\ d \end{pmatrix} = \begin{pmatrix} \var{-v1[0][0]}\\ \var{-v1[0][1]} \\ \var{-v1[0][2]} \end{pmatrix}$

$\begin{pmatrix} b \\ c \\ d \end{pmatrix} = \begin{pmatrix} \var{v2[0][0] -v1[0][0]}& \var{v3[0][0] -v1[0][0]} & \var{v4[0][0] -v1[0][0]}\\ \var{v2[0][1] -v1[0][1]}&\var{v3[0][1] -v1[0][1]} &\var{v4[0][1] -v1[0][1]} \\ \var{v2[0][2] -v1[0][2]}& \var{v3[0][2] -v1[0][2]}&\var{v4[0][2] -v1[0][2]} \end{pmatrix}^{-1} \begin{pmatrix} \var{-v1[0][0]}\\ \var{-v1[0][1]} \\ \var{-v1[0][2]} \end{pmatrix}=\begin{pmatrix} \var{c2} \\ \var{c3} \\ \var{c4} \end{pmatrix}$

Hence we obtain

$b=\var{c2}$ , $c=\var{c3}$ , $d=\var{c4}$ and hence $a=1- b-c-d= \var{c1}$ .

Substituting these values also correctly satisfy the 4th equation, so the equations are consistent and they have the unique solution given.

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
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet
Number entry
Number entry		Hasn't run yet

This question is used in the following exam:

Linear algebra by Simon Thomas in Maths support.

Loading...

Error

Simon's copy of Combine linearly dependent vectors to get zero

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

Loading...

Error

Simon's copy of Combine linearly dependent vectors to get zero

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Simon Thomas 5 years, 10 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`