Question

If a, b, c form a system of linear independent vectors then show that the system of vectors $ a-2b+c,2a-b+c\text{ and }3a+b+2c $ is also linearly independent.

Answer 1

Hint: In this question, we are given a, b, c as system of linear independent vector and we need to show $ a-2b+c,2a-b+c\text{ and }3a+b+2c $ as linear independent too. For this, we will try to write the given vectors (which we want to proof) in the form of linear combination with scalar $ {{x}_{1}},{{x}_{2}},{{x}_{3}} $ equating it to zero. Then we will rearrange them to make it a linear combination of a, b, c. Using the property of linear independence, we will put the scalar of a, b, c equal to zero and then prove $ {{x}_{1}}={{x}_{2}}={{x}_{3}}=0 $ which will give us required proof.

Complete step by step answer:
Here we are given that, a, b, c are linear independent vectors. According to definition of linear independent vectors for a, b, c if pa+qb+rc = 0 where p, q, r are scalar then p = q = r = 0.
Therefore, for the given vectors $ a-2b+c,2a-b+c\text{ and }3a+b+2c $ the linear combination with scalar $ {{x}_{1}},{{x}_{2}},{{x}_{3}} $ will be zero if $ {{x}_{1}}={{x}_{2}}={{x}_{3}} $ . Thus we can say that when $ {{x}_{1}}\left( a-2b+c \right)+{{x}_{2}}\left( 2a-b+c \right)+{{x}_{3}}\left( 3a+b+2c \right)=0 $ then we need to prove that $ {{x}_{1}}={{x}_{2}}={{x}_{3}}=0 $ .
We have equation as $ {{x}_{1}}\left( a-2b+c \right)+{{x}_{2}}\left( 2a-b+c \right)+{{x}_{3}}\left( 3a+b+2c \right)=0 $ .
Opening bracket, $ {{x}_{1}}a-2{{x}_{1}}b+{{x}_{1}}c+2{{x}_{2}}a-{{x}_{2}}b+{{x}_{2}}c+3{{x}_{3}}a+b{{x}_{3}}+2c{{x}_{3}}=0 $ .
Rearranging the terms we get, $ {{x}_{1}}a+2{{x}_{2}}a+3{{x}_{3}}a-2{{x}_{1}}b-{{x}_{2}}b+b{{x}_{3}}+{{x}_{1}}c+{{x}_{2}}c+2c{{x}_{3}}=0 $ .
Taking a common from first three terms, b common from the next three terms and c common from the last three terms we get, $ \left( {{x}_{1}}a+2{{x}_{2}}+3{{x}_{3}} \right)a+\left( -2{{x}_{1}}-{{x}_{2}}+{{x}_{3}} \right)b+\left( {{x}_{1}}+{{x}_{2}}+2{{x}_{3}} \right)c=0 $ .
Now we know that, a, b, c are linear independent vectors therefore their scalar must be equal to zero. So,
\[\begin{align}
  & {{x}_{1}}a+2{{x}_{2}}+3{{x}_{3}}=0\cdots \cdots \cdots \cdots \left( 1 \right) \\
 & -2{{x}_{1}}-{{x}_{2}}+{{x}_{3}}=0\cdots \cdots \cdots \cdots \left( 2 \right) \\
 & {{x}_{1}}+{{x}_{2}}+2{{x}_{3}}=0\cdots \cdots \cdots \cdots \left( 3 \right) \\
\end{align}\]
Adding equation (2) and (3) we get,
 $ -2{{x}_{1}}+{{x}_{1}}-{{x}_{2}}+{{x}_{2}}+{{x}_{3}}+2{{x}_{3}}=0\Rightarrow -{{x}_{1}}+3{{x}_{3}}=0\cdots \cdots \cdots \left( 4 \right) $ .
Multiplying equation (2) by 2 and adding with (1) we get,
\[\begin{align}
  & 2\left( -2{{x}_{1}}-{{x}_{2}}+{{x}_{3}} \right)+{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=0 \\
 & \Rightarrow -4{{x}_{1}}-2{{x}_{2}}+2{{x}_{3}}+{{x}_{1}}+2{{x}_{2}}+3{{x}_{3}}=0 \\
 & \Rightarrow -3{{x}_{1}}+5{{x}_{3}}=0\cdots \cdots \cdots \left( 5 \right) \\
\end{align}\]
Multiplying equation (4) by 3 and subtracting from (5) we get,
\[\begin{align}
  & -3{{x}_{1}}+5{{x}_{3}}-3\left( -{{x}_{1}}+3{{x}_{3}} \right)=0 \\
 & \Rightarrow -3{{x}_{1}}+5{{x}_{3}}+3{{x}_{1}}-9{{x}_{3}}=0 \\
 & \Rightarrow -4{{x}_{3}}=0 \\
 & \Rightarrow {{x}_{3}}=0 \\
\end{align}\].
Putting in (4), $ -{{x}_{1}}=0\Rightarrow {{x}_{1}}=0 $ .
Putting $ {{x}_{1}},{{x}_{2}} $ in (1) we get $ 3{{x}_{3}}=0\Rightarrow {{x}_{3}}=0 $ .
Therefore we see that $ {{x}_{1}}={{x}_{2}}={{x}_{3}} $ .
From our given equation if $ {{x}_{1}}\left( a-2b+c \right)+{{x}_{2}}\left( 2a-b+c \right)+{{x}_{3}}\left( 3a+b+2c \right)=0 $ then $ {{x}_{1}}={{x}_{2}}={{x}_{3}}=0 $ .
Thus given vector $ a-2b+c,2a-b+c\text{ and }3a+b+2c $ are linear independent vector.

Note:
 Students should keep in mind the way to prove linear dependency or independence of vectors. For solving the three equations, students can use any method. Take care of signs while rearranging the equation and subtracting/adding the equation.