2: Vectors
( \newcommand{\kernel}{\mathrm{null}\,}\)
After reading this chapter, you should be able to:
- define a vector
- add and subtract vectors,
- find linear combinations of vectors and their relationship to a set of equations,
- explain what it means to have a linearly independent set of vectors, and
- find the rank of a set of vectors.
What is a vector?
A vector is a collection of numbers in a definite order. If it is a collection of n numbers, it is called a n-dimensional vector. So, the vector →A given by
→A=[a1a2⋮an]
is a n-dimensional column vector with n components, a1,a2,......,an. The above is a column vector. A row vector [B] is of the form →B=[b1,b2,....,bn] where →B is a n-dimensional row vector with n components b1,b2,....,bn.
Give an example of a 3-dimensional column vector.
Solution
Assume a point in space is given by its (x,y,z) coordinates. Then if the value of x=3, y=2, z=5, the column vector corresponding to the location of the points is
[xyz]=[325]
When are two vectors equal?
Two vectors →A and →B are equal if they are of the same dimension and if their corresponding components are equal.
Given
→A=[a1a2⋮an]
and
→B=[b1b2⋮bn]
then →A=→B if ai=bi, i=1,2,......,n.
What are the values of the unknown components in →B if
→A=[2341]
and
→B=[b134b4]
and →A=→B.
Solution
b1=2,b4=1
How do you add two vectors?
Two vectors can be added only if they are of the same dimension and the addition is given by
[A]+[B]=[a1a2⋮an]+[b1b2⋮bn]=[a1+b1a2+b2⋮an+bn]
Add the two vectors
→A=[2341]
and
→B=[5−237]
Solution
→A+→B=[2341]+[5−237]=[2+53−24+31+7]=[7178]
A store sells three brands of tires: Tirestone, Michigan and Copper. In quarter 1, the sales are given by the column vector
→A1=[2556]
where the rows represent the three brands of tires sold – Tirestone, Michigan and Copper respectively. In quarter 2, the sales are given by
→A2=[20106]
What is the total sale of each brand of tire in the first half of the year?
Solution
The total sales would be given by
→C=→A1+→A2=[2556]+[20106]=[25+205+106+6]=[451512]
So, the number of Tirestone tires sold is 45, Michigan is 15 and Copper is 12 in the first half of the year.
What is a null vector?
A null vector (also called zero vector) is where all the components of the vector are zero.
Give an example of a null vector or zero vector.
Solution
The vector
[0000]
is an example of a zero or null vector.
What is a unit vector?
A unit vector →U is defined as
→U=[u1u2⋮un]
where
√u21+u22+u23+…+u2n=1
Give examples of 3-dimensional unit column vectors.
Solution
Examples include
[1√31√31√3],[100],[1√21√20],[010], etc.
How do youmultiply a vector by a scalar?
If k is a scalar and →A is a n-dimensional vector, then
k→A=k[a1a2⋮an]=[ka1ka2⋮kan]
What is 2→A if
→A=[25205]
Solution
2→A=2[25205]=[2×252×202×5]=[504010]
A store sells three brands of tires: Tirestone, Michigan and Copper. In quarter 1, the sales are given by the column vector
→A=[25256]
If the goal is to increase the sales of all tires by at least 25% in the next quarter, how many of each brand should be sold?
Solution
Since the goal is to increase the sales by 25%, one would multiply the →A vector by 1.25,
→B=1.25[25256]=[31.2531.257.5]
Since the number of tires must be an integer, we can say that the goal of sales is
→B=[32328]
What do you mean by a linear combination of vectors?
Given
→A1,→A2,......,→Am
as m vectors of same dimension n, and if k1,k2,...,km are scalars, then
k1→A1+k2→A2+.......+km→Am
is a linear combination of the m vectors.
Find the linear combinations
(a) →A−→B and
(b) →A+→B−3→C
where
→A=[236],→B=[112],→C=[1012]
Solution
(a) →A−→B=[236]−[112]=[2−13−16−2]=[124]
(b) →A+→B−3→C=[236]+[112]−3[1012]=[2+1−303+1−36+2−6]=[−2712]
What do you mean by vectors being linearly independent?
A set of vectors →A1,→A2,…,→Am are considered to be linearly independent if
k1→A1+k2→A2+.......+km→Am=→0
has only one solution of
k1=k2=......=km=0
Are the three vectors
→A1=[2564144], →A2=[5812], →A3=[111]
linearly independent?
Solution
Writing the linear combination of the three vectors
k1[2564144]+k2[5812]+k3[111]=[000]
gives
[25k1+5k2+k364k1+8k2+k3144k1+12k2+k3]=[000]
The above equations have only one solution, k1=k2=k3=0. However, how do we show that this is the only solution? This is shown below.
The above equations are
25k1+5k2+k3=0(1)
64k1+8k2+k3=0(2)
144k1+12k2+k3=0(3)
Subtracting Eqn (1) from Eqn (2) gives
39k1+3k2=0
k2=−13k1(4)
Multiplying Eqn (1) by 8 and subtracting it from Eqn (2) that is first multiplied by 5 gives
120k1−3k3=0
k3=40k1(5)
Remember we found Eqn (4) and Eqn (5) just from Eqns (1) and (2).
Substitution of Eqns (4) and (5) in Eqn (3) for k1 and k2 gives
144k1+12(−13k1)+40k1=0
28k1=0
k1=0
This means that k1 has to be zero, and coupled with (4) and (5), k2 and k3 are also zero. So the only solution is k1=k2=k3=0. The three vectors hence are linearly independent.
Are the three vectors
→A1=[125], →A2=[257], →A3=[61424]
linearly independent?
Solution
By inspection,
→A3=2→A1+2→A2
or
−2→A1−2→A2+→A3=→0
So the linear combination
k1→A1+k2→A2+k3→A3=→0
has a non-zero solution
k1=−2, k2=−2, k3=1
Hence, the set of vectors is linearly dependent.
What if I cannot prove by inspection, what do I do? Put the linear combination of three vectors equal to the zero vector,
k1[125]+k2[257]+k3[61424]=[000]
to give
k1+2k2+6k3=0(1)
2k1+5k2+14k3=0(2)
5k1+7k2+24k3=0(3)
Multiplying Eqn (1) by 2 and subtracting from Eqn (2) gives
k2+2k3=0
k2=−2k3(4)
Multiplying Eqn (1) by 2.5 and subtracting from Eqn (2) gives
−0.5k1−k3=0
k1=−2k3(5)
Remember we found Eqn (4) and Eqn (5) just from Eqns (1) and (2).
Substitute Eqn (4) and (5) in Eqn (3) for k1 and k2 gives
5(−2k3)+7(−2k3)+24k3=0
−10k3−14k3+24k3=0
0=0
This means any values satisfying Eqns (4) and (5) will satisfy Eqns (1), (2) and (3) simultaneously.
For example, chose
k3=6, then
k2=−12 from Eqn (4), and
k1=−12 from Eqn (5).
Hence we have a nontrivial solution of [k1k2k3]=[−12−126]. This implies the three given vectors are linearly dependent. Can you find another nontrivial solution?
What about the following three vectors?
[125],[257],[61425]
Are they linearly dependent or linearly independent?
Note that the only difference between this set of vectors and the previous one is the third entry in the third vector. Hence, equations (4) and (5) are still valid. What conclusion do you draw when you plug in equations (4) and (5) in the third equation: 5k1+7k2+25k3=0? What has changed?
Are the three vectors
→A1=[256489], →A2=[5813], →A3=[112]
linearly independent?
Solution
Writing the linear combination of the three vectors and equating to zero vector
k1[256489]+k2[5813]+k3[112]=[000]
gives
[25k1+5k2+k364k1+8k2+k389k1+13k2+2k3]=[000]
In addition to k1=k2=k3=0, one can find other solutions for which k1, k2, k3are not equal to zero. For example, k1=1, k2=−13, k3=40 is also a solution as
1[256489]−13[5813]+40[112]=[000]
Hence →A1, →A2, →A3 are linearly dependent.
What do you mean by the rank of a set of vectors?
From a set of n-dimensional vectors, the maximum number of linearly independent vectors in the set is called the rank of the set of vectors. Note that the rank of the vectors can never be greater than the vectors dimension.
What is the rank of
→A1=[2564144], →A2=[5812], →A3=[111]?
Solution
Since we found in Example 2.10 that →A1, →A2, →A3 are linearly independent, the rank of the set of vectors →A1, →A2, →A3 is 3. If we were given another vector →A4, the rank of the set of the vectors →A1, →A2, →A3,→A4 would still be 3 as the rank of a set of vectors is always less than or equal to the dimension of the vectors and that at least →A1, →A2, →A3 are linearly independent.
What is the rank of
→A1=[256489], →A2=[5813], →A3=[112]?
Solution
In Example 2.12, we found that →A1, →A2, →A3 are linearly dependent, the rank of →A1, →A2, →A3 is hence not 3, and is less than 3. Is it 2? Let us choose two of the three vectors
→A1=[256489], →A2=[5813]
Linear combination of →A1 and →A2 equal to zero has only one solution – the trivial solution. Therefore, the rank is 2.
What is the rank of
→A1=[112], →A2=[224], →A3=[335]?
Solution
From inspection,
→A2=2→A1,
that implies
2{\overrightarrow{A}}_{1} - {\overrightarrow{A}}_{2} + 0{\overrightarrow{A}}_{3} = \overrightarrow{0}\notag
Hence
k_{1}{\overrightarrow{A}}_{1} + k_{2}{\overrightarrow{A}}_{2} + k_{3}{\overrightarrow{A}}_{3} = \overrightarrow{0}\notag
has a nontrivial solution.
So {\overrightarrow{A}}_{1},\ \ {\overrightarrow{A}}_{2},\ \ {\overrightarrow{A}}_{3} are linearly dependent, and hence the rank of the three vectors is not 3. Since
{\overrightarrow{A}}_{2} = 2{\overrightarrow{A}}_{1},\notag
{\overrightarrow{A}}_{1}\text{ and }{\overrightarrow{A}}_{2} are linearly dependent, but
k_{1}{\overrightarrow{A}}_{1} + k_{3}{\overrightarrow{A}}_{3} = \overrightarrow{0}.\notag
has trivial solution as the only solution. So {\overrightarrow{A}}_{1} and {\overrightarrow{A}}_{3} are linearly independent. The rank of the above three vectors is 2.
Prove that if a set of vectors contains the null vector, the set of vectors is linearly dependent.
Let {\overrightarrow{A}}_{1},{\overrightarrow{A}}_{2},.........,{\overrightarrow{A}}_{m} be a set of n-dimensional vectors, then
k_{1}{\overrightarrow{A}}_{1} + k_{2}{\overrightarrow{A}}_{2} + \ \ldots\ + k_{m}{\overrightarrow{A}}_{m} = \overrightarrow{0}\notag
is a linear combination of the m vectors. Then assuming if {\overrightarrow{A}}_{1} is the zero or null vector, any value of k_{1} coupled with k_{2} = k_{3} = \ ..\ .\ = k_{m} = 0 will satisfy the above equation. Hence, the set of vectors is linearly dependent as more than one solution exists.
Prove that if a set of m vectors is linearly independent, then a subset of the m vectors also has to be linearly independent.
Let this subset of vectors be
{\overrightarrow{A}}_{a1},{\overrightarrow{A}}_{a2},\ldots,{\overrightarrow{A}}_{\text{ap}}\notag
where p < m.
Then if this subset of vectors is linearly dependent, the linear combination
k_{1}{\overrightarrow{A}}_{a1} + k_{2}{\overrightarrow{A}}_{a2} + \ldots + k_{p}{\overrightarrow{A}}_{\text{ap}} = \overrightarrow{0}\notag
has a non-trivial solution.
So
k_{1}{\overrightarrow{A}}_{a1} + k_{2}{\overrightarrow{A}}_{a2} + \ldots + k_{p}{\overrightarrow{A}}_{\text{ap}} + 0{\overrightarrow{A}}_{a(p + 1)} + ....... + 0{\overrightarrow{A}}_{\text{am}} = \overrightarrow{0}\notag
also has a non-trivial solution too, where {\overrightarrow{A}}_{a\left( p + 1 \right)},\ldots,{\overrightarrow{A}}_{\text{am}}are the rest of the (m - p) vectors. However, this is a contradiction. Therefore, a subset of linearly independent vectors cannot be linearly dependent.
Prove that if a set of vectors is linearly dependent, then at least one vector can be written as a linear combination of others.
Let {\overrightarrow{A}}_{1},{\overrightarrow{A}}_{2},\ldots,{\overrightarrow{A}}_{m} be linearly dependent set of vectors, then there exists a set of scalars
k_{1},\ldots,k_{m} not all of which are zero for the linear combination equation
k_{1}{\overrightarrow{A}}_{1} + k_{2}{\overrightarrow{A}}_{2} + \ldots + k_{m}{\overrightarrow{A}}_{m} = \overrightarrow{0}.\notag
Let k_{p} be one of the non-zero values of k_{i},\ i = 1,\ldots,m, that is, k_{p} \neq 0, then
A_{p} = - \frac{k_{2}}{k_{p}}{\overrightarrow{A}}_{2} - \ \ldots\ - \frac{k_{p - 1}}{k_{p}}{\overrightarrow{A}}_{p - 1} - \frac{k_{p + 1}}{k_{p}}{\overrightarrow{A}}_{p + 1} - \ \ldots\ - \frac{k_{m}}{k_{p}}{\overrightarrow{A}}_{m}\notag
and that proves the theorem.
Prove that if the dimension of a set of vectors is less than the number of vectors in the set, then the set of vectors is linearly dependent.
Can you prove it?
How can vectors be used to write simultaneous linear equations?
If a set of m simultaneous linear equations with n unknowns is written as
a_{11}x_{1} + \ \ldots\ + a_{1n}x_{n} = c_{1}\notag
a_{21}x_{1} + \ \ldots\ + a_{2n}x_{n} = c_{2}\notag
\begin{matrix} \vdots & & & \vdots \\ \vdots & & & \vdots \\ \end{matrix}\notag
a_{m1}x_{1} + \ \ldots\ + a_{\text{mn}}x_{n} = c_{n}\notag
where
x_{1},x_{2},\ldots,x_{n}are the unknowns, then in the vector notation they can be written as
x_{1}{\overrightarrow{A}}_{1} + x_{2}{\overrightarrow{A}}_{2} + \ldots + x_{n}{\overrightarrow{A}}_{n} = \overrightarrow{C}\notag
where
{\overrightarrow{A}}_{1} = \begin{bmatrix} a_{11} \\ \vdots \\ a_{m1} \\ \end{bmatrix}\notag
where
{\overrightarrow{A}}_{1} = \begin{bmatrix} a_{11} \\ \vdots \\ a_{m1} \\ \end{bmatrix}\notag
{\overrightarrow{A}}_{2} = \begin{bmatrix} a_{12} \\ \vdots \\ a_{m2} \\ \end{bmatrix}\notag
{\overrightarrow{A}}_{n} = \begin{bmatrix} a_{1n} \\ \vdots \\ a_{\text{mn}} \\ \end{bmatrix}\notag
{\overrightarrow{C}}_{1} = \begin{bmatrix} c_{1} \\ \vdots \\ c_{m} \\ \end{bmatrix}\notag
The problem now becomes whether you can find the scalars x_{1},x_{2},.....,x_{n} such that the linear combination
x_{1}{\overrightarrow{A}}_{1} + .......... + x_{n}{\overrightarrow{A}}_{n}\notag
is equal to the \overrightarrow{C}, that is
x_{1}{\overrightarrow{A}}_{1} + .......... + x_{n}{\overrightarrow{A}}_{n} = \overrightarrow{C}\notag
Write
25x_{1} + 5x_{2} + x_{3} = 106.8\notag
64x_{1} + 8x_{2} + x_{3} = 177.2\notag
144x_{1} + 12x_{2} + x_{3} = 279.2\notag
as a linear combination of set of vectors equal to another vector.
Solution
\begin{bmatrix} 25x_{1} & + 5x_{2} & + x_{3} \\ 64x_{1} & + 8x_{2} & + x_{3} \\ 144x_{1} & + 12x_{2} & + x_{3} \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}\notag
x_{1}\begin{bmatrix} 25 \\ 64 \\ 144 \\ \end{bmatrix} + x_{2}\begin{bmatrix} 5 \\ 8 \\ 12 \\ \end{bmatrix} + x_{3}\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 106.8 \\ 177.2 \\ 279.2 \\ \end{bmatrix}\notag
What is the definition of the dot product of two vectors?
Let \overrightarrow{A} = \left\lbrack a_{1},a_{2},\ldots,a_{n} \right\rbrack and \overrightarrow{B} = \left\lbrack b_{1},b_{2},\ldots,b_{n} \right\rbrack be two n-dimensional vectors. Then the dot product of the two vectors \overrightarrow{A} and \overrightarrow{B} is defined as
\overrightarrow{A} \cdot \overrightarrow{B} = a_{1}b_{1} + a_{2}b_{2} + \ldots + a_{n}b_{n} = \sum_{i = 1}^{n}{a_{i}b_{i}}\notag
A dot product is also called an inner product.
Find the dot product of the two vectors \overrightarrow{A} = [4, 1, 2, 3] and \overrightarrow{B} = [3, 1, 7, 2].
Solution
\begin{split} \overrightarrow{A} \cdot \overrightarrow{B} &= \lbrack 4,1,2,3\rbrack\ .\ \lbrack 3,1,7,2\rbrack\\ &= \left( 4 \right)\left( 3 \right) + \left( 1 \right)\left( 1 \right) + \left( 2 \right)\left( 7 \right) + \left( 3 \right)\left( 2 \right)\\ &= 33 \end{split}\notag
A product line needs three types of rubber as given in the table below.
Rubber type | Weight (lbs) | Cost per pound ($) |
---|---|---|
A B C |
200 250 310 |
20.23 30.56 29.12 |
Use the definition of a dot product to find the total price of the rubber needed.
Solution
The weight vector is given by
\overrightarrow{W} = \lbrack 200,250,310\rbrack\notag
and the cost vector is given by
\overrightarrow{C} = \lbrack 20.23,30.56,29.12\rbrack\notag
The total cost of the rubber would be the dot product of \overrightarrow{W} and \overrightarrow{C}.
\begin{split} \overrightarrow{W} \cdot \overrightarrow{C} &= \lbrack 200,250,310\rbrack \cdot \lbrack 20.23,30.56,29.12\rbrack\\ &= (200)(20.23) + (250)(30.56) + (310)(29.12)\\ &= 4046 + 7640 + 9027.2\\ &= \text{\$} 20713.20 \end{split}\notag
Vectors Quiz
A set of equations
4x_{1} + 7x_{2} + 11x_{3} = 13\notag
17x_{1} + 39x_{2} + 23x_{3} = 31\notag
13x_{1} + 67x_{2} + 59x_{3} = 37\notag
can also be written as
(A) x_{1}\begin{bmatrix} 4 \\ 17 \\ 13 \\ \end{bmatrix} + x_{2}\begin{bmatrix} 7 \\ 39 \\ 23 \\ \end{bmatrix} + x_{3}\begin{bmatrix} 11 \\ 23 \\ 59 \\ \end{bmatrix} = \begin{bmatrix} 13 \\ 31 \\ 37 \\ \end{bmatrix}
(B) 4\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} + 39\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} + 59\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ \end{bmatrix} = \begin{bmatrix} 13 \\ 31 \\ 37 \\ \end{bmatrix}
(C) x_{1}\begin{bmatrix} 4 \\ 7 \\ 11 \\ \end{bmatrix} + x_{2}\begin{bmatrix} 17 \\ 39 \\ 23 \\ \end{bmatrix} + x_{3}\begin{bmatrix} 13 \\ 67 \\ 59 \\ \end{bmatrix} = \begin{bmatrix} 13 \\ 31 \\ 37 \\ \end{bmatrix}
(D) x_{1}\begin{bmatrix} 13 \\ 17 \\ 4 \\ \end{bmatrix} + x_{2}\begin{bmatrix} 67 \\ 39 \\ 7 \\ \end{bmatrix} + x_{3}\begin{bmatrix} 59 \\ 23 \\ 11 \\ \end{bmatrix} = \begin{bmatrix} 57 \\ 13 \\ 31 \\ \end{bmatrix}
The magnitude of the vector, V = (5, - 3,2) is
(A) 4
(B) 10
(C) \sqrt{38}
(D) \sqrt{20}
The rank of the vector
\overset{\rightarrow}{A}\begin{bmatrix} 2 \\ 3 \\ 7 \\ \end{bmatrix},\begin{bmatrix} 6 \\ 9 \\ 21 \\ \end{bmatrix},\begin{bmatrix} 3 \\ 2 \\ 7 \\ \end{bmatrix}\notag
is
(A) 1
(B) 2
(C) 3
(D) 4
If \overrightarrow{A} = (5,2,3)and \overrightarrow{B} = (6, - 7,3), then 4\overrightarrow{A} + 5\overrightarrow{B}is
(A) (50, - 5,6)
(B) (50, - 27,27)
(C) (11, - 5,6)
(D) (20,8,12)
The dot product of two vectors \overset{\rightarrow}{A} and \overset{\rightarrow}{B}
\overset{\rightarrow}{A} = 3i + 5j + 7k\notag
\overset{\rightarrow}{B} = 11i + 13j + 17k\notag
most nearly is
(A) 14.80
(B) 33.00
(C) 56.00
(D) 217.0
The angle in degrees between two vectors \overrightarrow{u} and \overrightarrow{v}
\overset{\rightarrow}{u} = 3i + 5j + 7k\notag
\overset{\rightarrow}{v} = 11i + 13j + 17k\notag
most nearly is
(A) 8.124
(B) 11.47
(C) 78.52
(D) 81.88
Vectors Exercise
For
\overrightarrow{A} = \begin{bmatrix} 2 \\ 9 \\ - 7 \\ \end{bmatrix},\overrightarrow{B} = \begin{bmatrix} 3 \\ 2 \\ 5 \\ \end{bmatrix},\overrightarrow{C} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}
find \overrightarrow{A} + \overrightarrow{B} and 2\overrightarrow{A} - 3\overrightarrow{B} + \overrightarrow{C}.
- Answer
-
\begin{bmatrix} 5 \\ 11 \\ - 2 \\ \end{bmatrix}; \begin{bmatrix} - 4 \\ 13 \\ - 28 \\ \end{bmatrix}
Are
\overrightarrow{A} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix},\overrightarrow{B} = \begin{bmatrix} 1 \\ 2 \\ 5 \\ \end{bmatrix},\overrightarrow{C} = \begin{bmatrix} 1 \\ 4 \\ 25 \\ \end{bmatrix}
linearly independent?.
What is the rank of the above set of vectors?
- Answer
-
3
Add exercises text here.
Are
\overrightarrow{A} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix},\overrightarrow{B} = \begin{bmatrix} 1 \\ 2 \\ 5 \\ \end{bmatrix},\overrightarrow{C} = \begin{bmatrix} 3 \\ 5 \\ 7 \\ \end{bmatrix}
linearly independent?.
What is the rank of the above set of vectors?
- Answer
-
3
Are
\overrightarrow{A} = \begin{bmatrix} 1 \\ 2 \\ 5 \\ \end{bmatrix},\overrightarrow{B} = \begin{bmatrix} 2 \\ 4 \\ 10 \\ \end{bmatrix},\overrightarrow{C} = \begin{bmatrix} 1.1 \\ 2.2 \\ 5.5 \\ \end{bmatrix}
linearly independent?
What is the rank of the above set of vectors?
- Answer
-
No;1
If a set of vectors contains the null vector, the set of vectors is linearly
- Independent
- Dependent?
- Answer
-
B
If a set of vectors is linearly independent, a subset of the vectors is linearly
- Independent.
- Dependent.
- Answer
-
A
If a set of vectors is linearly dependent, then
- At least one vector can be written as a linear combination of others.
- At least one vector is a null vector.
- Answer
-
A
If the dimension of a set of vectors is less than the number of vectors in the set, then the set of vectors is linearly
- Dependent.
- Independent.
- Answer
-
A
Find the dot product of\overrightarrow{A} = (2,1,2.5,3)and \overrightarrow{B} = ( - 3,2,1,2.5)
- Answer
-
6
If \overrightarrow{u},\overrightarrow{v},\overrightarrow{w} are three nonzero vector of 2-dimensions, then
- \overrightarrow{u},\overrightarrow{v},\overrightarrow{w} are linearly independent
- \overrightarrow{u},\overrightarrow{v},\overrightarrow{w} are linearly dependent
- \overrightarrow{u},\overrightarrow{v},\overrightarrow{w} are unit vectors
- k_{1}\overrightarrow{u} + k_{2}\overrightarrow{v} + k_{3}\overrightarrow{v} = \overrightarrow{0} has a unique solution.
- Answer
-
B
\overrightarrow{u}and \overrightarrow{v}are two non-zero vectors of dimension n. Prove that if \overrightarrow{u}and \overrightarrow{v}are linearly dependent, there is a scalar q such that \overrightarrow{v} = q\overrightarrow{u}.
- Answer
-
Hint :
Start with k_{1}\overrightarrow{u} + k_{2}\overrightarrow{v} = \overrightarrow{0}
Show that k_{1} \neq 0and k_{2} \neq 0 because \overrightarrow{u}and \overrightarrow{v} are both nonzero.
Hence
\begin{split} \overrightarrow{\nu} &= - \frac{k_{1}}{k_{2}}\overrightarrow{u}\\ &=q\overrightarrow{u} \;\;\;\;\;\;\;\ q=- \frac{k_{1}}{k_{2}} \end{split}
\overrightarrow{u}and \overrightarrow{v}are two non-zero vectors of dimension n. Prove that if there is a scalar q such that \overrightarrow{v} = q\overrightarrow{u}, then \overrightarrow{u}and \overrightarrow{v}are linearly dependent.
- Answer
-
Hint:
Since
\begin{matrix} \overrightarrow{v} = q\overrightarrow{u} \\ \overrightarrow{v} - q\overrightarrow{u} = \overrightarrow{0} \\ \end{matrix}
q \neq 0, otherwise \overrightarrow{v} = \overrightarrow{0}
So the equation
k_{1}\overrightarrow{v} + k_{2}\overrightarrow{u} = \overrightarrow{0}\notag
has a non trivial solution of
k_{1} = 1,k_{2} = q \neq 0.\notag