Determine whether the given vectors from linearly dependent set or linearly independent set.

Question:

1) Show that u= \(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\) can be expressed as a linear combination of

u1= \(\begin{bmatrix} 4 \\0  \\2 \end{bmatrix}\) u2=\(\begin{bmatrix}  0 \\5  \\ 0 \end{bmatrix}\) u3=\(\begin{bmatrix} 2 \\0  \\6  \end{bmatrix}\)

2) Determine whether the vectors given from a linearly dependent set or linearly independent set

a) u1= \(\begin{bmatrix} 2 \\2  \\0 \end{bmatrix}\)   u2= \(\begin{bmatrix}  2 \\ -2 \\ 0 \end{bmatrix}\)   u3= \(\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}\)

 

b) u1= \(\begin{bmatrix} 3 \\-1  \\-1 \\2 \end{bmatrix}\) u2= \(\begin{bmatrix} 1 \\0  \\2 \\1 \end{bmatrix}\) u3= \(\begin{bmatrix} 3 \\-1  \\0 \\1 \end{bmatrix}\)

Summary:

1) Yes, the variables u1, u2, and u3 can be used to denote
u=(-3/10) u1+(0) u2+(11/10)u3
2)
a) A linearly independent set is formed by the vectors u1, u2, and u3.
b) A linearly independent set is formed by the vectors u1, u2, and u3.
Explanation:
Given vector u=\(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\) has to be expressed
linear combination of u1= \(\begin{bmatrix} 4 \\0  \\2 \end{bmatrix}\) u2= \(\begin{bmatrix}  0 \\5  \\ 0 \end{bmatrix}\) u3= \(\begin{bmatrix} 2 \\0  \\6  \end{bmatrix}\)
Let u= au1 + bu2 +cu3
where a,b,c be some unknown constants, which expresses u as linear
combination of u1,u2u3.

a=\(\begin{bmatrix} 4 \\0  \\2 \end{bmatrix}\) + b= \(\begin{bmatrix}  0 \\5  \\ 0 \end{bmatrix}\) + c=

\(\begin{bmatrix} 2 \\0  \\6  \end{bmatrix}\) = \(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\)

\(\begin{bmatrix} 4a \\0  \\2a \end{bmatrix}\) + b= \(\begin{bmatrix}  0 \\5b  \\ 0 \end{bmatrix}\) + c=

\(\begin{bmatrix} 2c \\0  \\6c \end{bmatrix}\) = \(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\)

\(\begin{bmatrix} 4a &0  &2a \\ 0& 5b& 0 \\ 2c &0 &6c \end{bmatrix}\) = \(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\)

Equivalize each element of both left and right matrices.
4a + 2C = 1 =) (4a=1-2C) =(2a=1/2-c)        —— eq (2)
5b = 0 b=0          ——- eq (1)
2a + 6C = 6         ——- eq (3)
Substituting 2 in 3
1/2-c+6c-¿5c+1/2=6
5c=11/2
c=11/10         ——- eq (4)
Substituting 4 in equation 2
4a = 1 – 2C
4a = 1 – 2(11/10) =) 4a = 1 – 11/5
4a=5-11/5=-6/5
a=-6/5*1/2=-3/10     ——- eq (5)
From equation 1,4, 5
 \(\begin{bmatrix} 1 \\0  \\6  \end{bmatrix}\)=-3/10 \(\begin{bmatrix} 4 \\0  \\2 \end{bmatrix}\) +0 \(\begin{bmatrix}  0 \\5  \\ 0 \end{bmatrix}\) +11/10 \(\begin{bmatrix} 2 \\0  \\6  \end{bmatrix}\)
u = (- 3/10) u1 + 0u2 + 11/10 u3.
b) Given set of vectors are:

u1= \(\begin{bmatrix} 3 \\-1  \\-1 \\2 \end{bmatrix}\) u2= \(\begin{bmatrix} 1 \\0  \\2 \\1 \end{bmatrix}\) u3= \(\begin{bmatrix} 3 \\-1  \\0 \\1 \end{bmatrix}\)

we have to find if the vectors are linearly independent or dependent.
M= \(\begin{bmatrix} 3 & 1& 3 & 0 \\-1 & 0 & -1& 0  \\-1 & 2 & 0 & 0\\2 &1 &1 &0 \end{bmatrix}\)
we have to bring it to the G@@5; triangular matrix by using now, transformations.
R->3Rq+R
R3 ->3R3+R
R4 ->3R4-2R1
M= \(\begin{bmatrix} 3 & 1& 3 & 0 \\0 & 1 & 0 & 0  \\0 & 7 & 3 & 0\\0 &1 &-3 &0 \end{bmatrix}\)
Using now transformation R3->R3-7R R4->R4-R2
M= \(\begin{bmatrix} 3 & 1& 3 & 0 \\0 & 1 & 0 & 0  \\0 & 0 & 3 & 0\\0 &1 &-3 &0 \end{bmatrix}\)
Using transformation
R4->R4+R3
Rank = no of non-zero rows = 3 no. of vectors
The given vector form a linearly independent set.

 

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