Linear Combination: Example Solved Problems

LINEAR COMBINATION

WORKED EXAMPLE PROBLEMS

Example 1

Let V=R², S={(1, 2), (2, 1) }, v = (2, 2) ∈ V and α, β ∈ F. Check whether v is a linear combination of v₁ and v₂ (ie) v = αv₁ + βv₂.

Solution:

Given that v = αv₁ + βv₂.

Let(2, 2) = α(1, 2) + β(2, 1) = (α, 2α) + (2β, β)

= (α + 2β, 2α + β)

Comparing on both sides we get

 β=2/3

Then α +2β = 2 becomes

 α+2(2/3) = 2

 α = 2 – 4/3 = 2/3

 α = 2/3

 The values of α and β are α=2/3 and β=2/3.

Substitute the values of α and β in the ordered pair we get,

  (2, 2) = α(1, 2) + β(2, 1)

 = 2/3(1,2) + 2/3(2,1)

= (2, 2)

v = αv₁ + βv₂

 Hence v is a linear combination of v₁ and v₂.

Example 2

Let V=R³, S = {(1, 2, 0), (0, −5, −7) } and v = (2,−5,7) ∈ V. Verify whether v is a linear combination of S or not.

Solution:

Let us consider v = αv₁ + βv₂

 (2,−5,7) = α(1, 2, 0) + β(0, −5,−7) = (α, 2α, 0) + (0, −5β, −7 β)

 (2,−5,7) = (α, 2α−5β, −7B)

Comparing on both sides we get α = 2, 2α−5β = −5 and −7β=7

From the first and last terms, we have

α=2 and β=− 1.

Substitute these two values in the middle term, we get

2(2)−5(−1)=−5

⇒ 9 ≠ −5

v is not a linear combination of S.

Example 3

Let v₁ = (2, 1), and v₂ = (−4,−2). Verify whether x=(3,5) is a linear combination of v₁ and v₂.

Solution:

Let us consider α, β ∈ F such that αv₁ + βv₂ = x.

 α(2, 1) + β(−4,−2) = (3, 5)

(2α, α)+(−4β, 2β)=(3,5)

(2α−4β, α−2β) = (3, 5)

Comparing on both sides, 2α−4β=3 and

 α−2β=5

 = −4+4 = 0.

 Δ=0, this system is inconsistent. Hence there is no solution, and x is not a linear combination of v₁ and v₂.

Example 4

If v = (−2, 0, 3) is a linear combination of (1, 3, 0) and (2, 4, −1) in R³. Prove that it is a linear combination of (1, 3, 0) and (2, 4, − 1).

Solution:

We know that v=αv₁+βv₂.

Let (−2, 0,3) = α(1, 3, 0) + β(2, 4, ‒1) = (α, 3α, 0) + (2β, 4β, −β)

(−2, 0, 3) = (α + 2β, 3α + 4β, −β)

Comparing on bothsides, we have

⇒ α+2β=−2

3α +4β=0

−β=3

 β=−3.

α+2(−3)=−20

α−6=−2

α= 4.

3(4)+4(−3)=0

Hence (−2, 0, 3) is a linear combination of (1, 3, 0) and (2, 4, −1).

Example 5

Write the vector v = (1, ‒2, 5) as a linear combination of the vectors e₁ = (1, 1, 1); e₂ = (1, 2, 3);  e₃ = (2, − 1, 1).

Solution:

We know that v = ae₁+ be₂ + ce₃

(we know that v = αv₁+ βv₂)

(1, −2, 5) = a(1, 1, 1) + b (1, 2, 3) + c (2, − 1, 1)

 = 1(2+3) − 1(1 + 1) + 2(3−2) = 5−2+2 = 5

 a= Δa / Δ = − 30/5 = ‒6

 b= Δb / Δ = 15/5 = 3

 c= Δc / Δ = − 10/5 = 2

  v = −6e₁ + 3e₂ + 2e₃.

Example 6

Write the vectors v = (2,−5, 3) as a linear combination of the vectors e₁=(1, − 3, 2), e₂=(2,−4, −1), e₃=(1,−5, 7)

Solution:

Let (2,−5,3) = a(1, −3, 2) + b (2,−4,−1) + c(1, −5, 7)

  = 1[−28+(−5)]−2 [−21+10] +1 (3+8)

= −33+22+11=0

The system is inconsistent and so has no solution.

  v is not a linear combination of e₁, e₂, e₃.

Example 7

For which values of k will the vector v = (1, − 2, k) in R³ be a linear combination of the vectors u = (3, 0,−2) and w = (2,−1,−5).

Solution:

Let (1,−2, k) = α(3, 0, −2) + β(2,−1,−5)

= (3α, 0, −2α) + (2β, − β,− 5β)

(1, −2, k) = (3α+2β, −β, −2α−5β)

Comparing the coefficients on both sides we get

 3α + 2β = 1              ….(1)

 β = 2              ….(2)

− 2α−5β= k              ….(3)

Substitute equation (2) in equation (1) we get

3α = 1−2β = 1 − 2(2)

⇒ 3α = −3

 α = − 1.

Now

 −2α−5β = k becomes

−2(−1)−5(2) = k,   (α=−1, β=2)

2+(−10)=k

 k =−8

Example 8

Write the matrix E =  as a linear combination of the matrix .

Solution:

Let E = αA+βB+γC

Equating the corresponding numbers we have

 α=3; α+2γ=1, α+β= 1 and β−γ=−1

 α = 3

α+2γ=1

2γ=1−3

γ=−1

α+β=1

3+β=1

β=−2

β− γ =−1

 −2+1=−1

 E=3A−2B−C

Example 9:

In R³, verify whether the first vector can be expressed as a linear combination of the other two or not, for (3, 4, 1), (1, − 2, 1), (− 2, − 1, 1).

Solution:

We know that v = αv₁ +βv₂

Let v = (3, 4, 1), v₁ = (1,−2, 1) and v₂ = (−2, −1, 1)

  (3, 4, 1) = α(1, −2, 1) + β(−2, −1, 1)

 = (α, −2α, α) + (−2β, −β, β)

(3, 4, 1) = (α−2β, −2α−β, α+ β)

Comparing on both sides, we have

 α−2β=3, −2α− β = 4 and α + β = 1

Now − 2α−β = 4

 α+β =1

Then

 −α=5

 α=−5

⇒ α+ β =1⇒

−5+ β =1

⇒ β = 6

 α− 2β = 3

⇒ − 5 −2 (6) = − 17 ≠ 3.

Hence v cannot be expressed as a linear combinations of other two.

Example 10

Verify x³−3x+5 is a linear combination of x³+2x²−x+1 and x³+3x²−1.

Solution:

Let v=av₁+bv₂

  x³ − 3x + 5 = a (x³ + 2x² −x + 1) + b(x³ + 3x² − 1)

 = ax³ +2ax² − ax + a + bx³ + 3bx² − b

  x³ − 3x + 5 = x³(a + b) + x²(2a + 3b) + x(− a) + (a − b).

Equating the coefficients on both sides, then we get

a+b=1; 2a+3b= 0; − a=−3 and a−b=5.

From the last two equations we have a = 3 and b=−2.

  a+b=1

⇒ 3−2=1 holds good.

2a+3b=0

⇒2 (3) + 3 (− 2) = 0 holds good.

  All the equations are satisfied.

Hence x³−3x+5 is a linear combination of x³ + 2x²−x+1 and x³+3x²−1.

Example 11

Verify whether the first polynomial can be expressed as a linear combination of the other two in P₃(R) for the given x³−8x²+4x, x³−2x²+3x−1 and x³−2x+3.

Solution:

Let v=av₁+bv₂

 x³− 8x²+4x= a (x³ − 2x² + 3x − 1) + b (x³ − 2x + 3)

 x³−8x²+4x= ax³−2ax²+3ax − a + bx³ − 2bx+ 3b

 = x³(a+b) + x²(−2a) + x(3a−2b) + (−a+3b)

Equating the coefficients on both sides we get

 1=a+b, −8=−2a, 4=3a−2b & −a+3b=0

a+b=1       ...(1)

2a = 8         ...(2)

3a−2b=4       ...(3)

−a+3b=0        ...(4)

From equation (2): a = 8/2 = 4        ...(5)

 b = 1−a = 1−4

 b = −3

 a = 4 & b = −3.

Substitute the values of a & b in equation (3) we get

 3(4) − 2(− 3) = 12+6 = 18 ≠ 4

Substitute the values of a & b in equation (4) we get

 − (4) + 3(− 3) = −4−9=−13 ≠ 0

For the values a=4 & b=−3 equations (3) and (4) are not satisfied.

 x³− 8x² + 4x is not a linear combination of

 x³− 2x² + 3x − 1 & x³ − 2x + 3.