It made possible to stone food and use it over a longer period. ( 5std answer name the following)
Answers
Answer:Find a basis for the subspace of R4 spanned by the given vectors.
(1,1,-5,-6), (2,0,2,-2), (3,-1,0,8).
2. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.
v1=(1,0,1,1), v2 = (-4,4,-1,4), v3 = (2,4,5,10), v4 = (-10,4,-7,-2)
3. Find a subset of the vectors that forms a basis for the space spanned by the vectors, then express each vector that is not in the basis as a linear combination of the basis vectors.
v1=(1,-1,5,2), v2 = (-2,3,1,0), v3 = (5,-6,14,6), v4 = (0,4,2,-3), v5 = (3,15,45,2)
4. Determine whether b is in the column space of A, and if so, express b as a linear combination of the column vectors of A.
A = 3 4 B = -1
6 -9 15
Expert's answer
1.As the set { v_1,v_2,v_3v
1
,v
2
,v
3
} span a subspace of R^{4}R
4
then, c_1v_1+c_2v_2+c_3v_3=0c
1
v
1
+c
2
v
2
+c
3
v
3
=0
c_1(1,1,-5,-6)+c_2(2,0,2,-2)+c_3(3,-1,0,8)=0c
1
(1,1,−5,−6)+c
2
(2,0,2,−2)+c
3
(3,−1,0,8)=0
We get following equations from above equation
c_1+2c_2+3c_3=0c
1
+2c
2
+3c
3
=0
c_1-c_3=0c
1
−c
3
=0
-5c_1+2c_2=0−5c
1
+2c
2
=0
-6c_1-2c_2+8c_3=0−6c
1
−2c
2
+8c
3
=0
From the above equation we can easily conclude that c_1=c_2=c_3=0c
1
=c
2
=c
3
=0
So,{ v_1,v_2,v_3v
1
,v
2
,v
3
} are linearly independent.
Thus, the set is the basis for the subspace of R4.
2.As the set v_1,v_2,v_3,v_4v
1
,v
2
,v
3
,v
4
span R4.
then,c_1v_1+c_2v_2+c_3v_3+c_4v_4=0c
1
v
1
+c
2
v
2
+c
3
v
3
+c
4
v
4
=0
c_1(1,0,1,1)+c_2(-4,4,-1,4)+c_3(2,4,5,10)+c_4(-10,4,-7,-2)=0c
1
(1,0,1,1)+c
2
(−4,4,−1,4)+c
3
(2,4,5,10)+c
4
(−10,4,−7,−2)=0
We get equations:
c_1 -4 c_2 +2 c_3 -10 c_4 = 0c
1
−4c
2
+2c
3
−10c
4
=0
4 c_2 +4 c_3 +4 c_4 = 04c
2
+4c
3
+4c
4
=0
c_1 -c_2 +5 c_3 -7 c_4 = 0c
1
−c
2
+5c
3
−7c
4
=0
c_1 +4 c_2 +10 c_3 -2 c_4 = 0c
1
+4c
2
+10c
3
−2c
4
=0
Solving these equations we can easily get
c_1 = -6 c_3 +6 c_4c
1
=−6c
3
+6c
4
c_2 = -c_3 -c_4c
2
=−c
3
−c
4
c_3 = arbitraryc
3
=arbitrary
c_4 = arbitraryc
4
=arbitrary
{ v_1,v_2v
1
,v
2
} span { v_3,v_4v
3
,v
4
}
{v_1,v_2v
1
,v
2
} spans a subspace of R4 and vectors are linearly independent.Thus,vectors serve as the basis of a subspace of R4.
3.c_1v_1+c_2v_2+c_3v_3+c_4v_4+c_5v_5=0c
1
v
1
+c
2
v
2
+c
3
v
3
+c
4
v
4
+c
5
v
5
=0
c_1(1,-1,5,2)+c_2(-2,3,1,0)+c_3(5,-6,14,6)+c_4(0,4,2,-3)+c_5(3,15,45,2)=0c
1
(1,−1,5,2)+c
2
(−2,3,1,0)+c
3
(5,−6,14,6)+c
4
(0,4,2,−3)+c
5
(3,15,45,2)=0
We get the following equations from the above equation.
c_1 -2 c_2 +5 c_3 +3 c_5 = 0c
1
−2c
2
+5c
3
+3c
5
=0
- c_1 +3 c_2 -6 c_3 +4 c_4 +15 c_5 = 0−c
1
+3c
2
−6c
3
+4c
4
+15c
5
=0
5 c_1 +c_2 +14 c_3 +2 c_4 +45 c_5 = 05c
1
+c
2
+14c
3
+2c
4
+45c
5
=0
2 c_1+6 c_3 -3 c_4 +2 c_5 = 02c
1
+6c
3
−3c
4
+2c
5
=0
Solving these equations we can get:
c_1 = -3 c_3 -7 c_5c
1
=−3c
3
−7c
5
c_2 =c_3 -2 c_5c
2
=c
3
−2c
5
c_3 = arbitraryc
3
=arbitrary
c_4 = -4 c_5c
4
=−4c
5
c_5 = arbitraryc
5
=arbitrary
{v_1,v_2,v_4v
1
,v
2
,v
4
} span {v_3,v_5v
3
,v
5
}
{ v_1,v_2,v_4v
1
,v
2
,v
4
} span a subspace of R4 and vectors are linearly independent.Thus, vectors serve as the basis of a subspace of R4.
4.A=\begin{bmatrix} 3 & 4 \\ 6 & -9 \end{bmatrix}A=[
3
6
4
−9
]
B=\begin{bmatrix} -1 \\ 15 \end{bmatrix}B=[
−1
15
]
B=c_1\begin{pmatrix} 3 \\ 6 \end{pmatrix}+c_2\begin{pmatrix} 4 \\ -9 \end{pmatrix}B=c
1
(
3
6
)+c
2
(
4
−9
)
3c_1+4c_2=-13c
1
+4c
2
=−1
6c_1-9c_2=156c
1
−9c
2
=15
c_1=1,c_2=-1c
1
=1,c
2
=−1
Yes, B is in the column space of A as it can be expressed as a linear combination of column vectors of A.
Explanation: