Combine linearly dependent vectors to get zero

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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

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Question statement

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

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

You are asked to show that the vectors $\textbf{v}_1,\;\textbf{v}_2,\;\textbf{v}_3,\;\textbf{v}_4$ are linearly dependent.

Find 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([-3,6,-6,0])
c	integer	3
b	integer	-3
ga	integer	-1
al	integer	-1
y	matrix	matrix([-6,3,0,-6])
v1	matrix	matrix([-3,6,-6,0])
v2	matrix	matrix([-6,3,0,-6])
v3	matrix	matrix([0,0,-3,0])
v4	matrix	matrix([-3,-3,9,-6])
t	integer	4
w	matrix	matrix([-3,-3,9,-6])
c2	rational	-1/2
c3	rational	1/2
del	integer	-1
c1	rational	1/2
z	matrix	matrix([0,0,-3,0])
c4	rational	1/2

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Generated value: `integer`

-3

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

Find $a,\;b,\;c$ and $d$

$a=$ 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:

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

On solving these equations we see that although there are more equations than unknowns, the equations are consistent and they have the solution:

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

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This question is used in the following exams:

Concrete vector spaces by Philip Walker in MATH1060.
Linear algebra by Newcastle University Mathematics and Statistics in Content created by Newcastle University.
Linear algebra by Henrik Skov Midtiby in SDS.
Maria's copy of Linear algebra by Maria Aneiros in Maria's workspace.
Ann's copy of Linear algebra by Ann Smith in Linear Algebra.
Linear Algebra 1 by Ugur Efem in Ugur's workspace.
Farai's copy of Linear algebra by Farai Nechikwira in Farai's workspace.
Farai's copy of Farai's copy of Linear algebra by Farai Nechikwira in Farai's workspace.

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Combine linearly dependent vectors to get zero

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Ugur Efem 5 months, 4 weeks ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Combine linearly dependent vectors to get zero

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Taxonomy: mathcentre

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Ugur Efem 5 months, 4 weeks ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: integer

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Generated value: `integer`