Answer to Question #106621 in Linear Algebra for Sourav mondal

Question #106621
1.Let B = (a1,a2, a3) be an ordered basis of
R3 with al= (1, 0, -1), a2= (1, 1, 1),
a3= (1, 0, 0). Write the vector v = (a, b, c) as
a linear combination of the basis vectors
from B.

2.Suppose al= (1, 0, 1), a2= (0, 1, -2) and
a3 = (-1, -1, 0) are vectors in R3and
f : R3 -> R is a linear functional such that
f(al) = 1, f(a2) = -1 and f(a3) = 3. If
a = (a, b, c) E R3, find f(a).
1
Expert's answer
2020-03-26T12:27:08-0400

We need to write the vector ""v"" linear combination of the basis vectors of B

From the given data

"a_1 = (1,0,1), \\space a_2 = (1, 1, 1) \\space and \\space a_3 =(1, 0, 0)"


Now we can write the vector "v" as linear combination of the vectors "a_1, a_2 \\space and \\space a_3"



"v = xa_1 + ya_2 + za_3"

"(a, b, c)= x(1,0,1) + y (1,1,1) + z (1, 0, 0)"

"(a, b, c)= (x,0,x) + (y, y, y) + (z, 0, 0)"

"(a, b, c)= (x+ y + z, y ,x+ y)"

compare both sides ,

"x + y + z = a\\\\\ny=b \\\\\nx + y = c"

Plug "y = b" in the equation "x + y = c" then "x = c - y = c - b"

Now

"x + y + z =a"

"c - b + b + z = a"

"z = a - c"

Now we can write the linear combination as

"(a, b, c)= (c-b)( 1,0,1) + b (1,1,1) + (a-c) (1, 0, 0)"


2).

"a_1 =(1, 0, 1), a_2 = (0, 1, 2) \\space and \\space a_3 = (-1, -1, 0)"

Given, f is a linear function


and "f (a_1) = 1, f(a_2) = -1 , f(a_3) = 3"


Let the function f(x) = ax+by + c z


"f (a_1) = 1" It means,


"f (1, 0, 1) =1"


"a(1) + b(0) +c(1) =1 \\\\\na + c =1 ................................(A)"


"f(a_2) = -1"


"f(0, 1, 2) = -1\\\\\na(0) + b(1) +c(-2) =-1\\\\\nb -2 c = -1 .................................(B)"


"f(a_3) = 3 \\\\\nf(-1, -1, 0) = 3\\\\\na(-1) +b(-1) +c(0) = 3\\\\\n-a -b = 3 .............................(C)"


From the equations A, B and C

"c = 1- a \\space"

Plug this c in the equation B, then

"b -2 + 2a =- 1 \\\\\nb + 2a = 1 .....................(D)"

Equation (C) is "- a - b = 3"


From equation C and D


"b + 2a - a - b = 1 + 3 \\\\\na =4"

"c = 1- a = 1- 4 = - 3"

"b - 2c = -1 \\\\\n\nb - 2(-3) = -1 \\\\\nb + 6 = -1\\\\\nb = -7"

So, the answer is

"(a, b, c) = ( 4, -7, - 3 )"

"f(a) = f(a, b, c) = f(4, -7, -3) = 4x - 7y -3z"


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