Question

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

Answer 1

Given a, b, c are linearly independent
$\therefore pa+qb+rc=0\Rightarrow p=0,q=0,r=0$...(1)
Now consider $x(a-2b+c)+y(2a-b+c)+z(3a+b+2c)=0$ or
$(x+2y+3z)a+(-2x-y+z)b+(x+y+2z)c=0$
Hence by (1) we have
$x+2y+3z=0,-2x-y+z=0,x+y+2z=0$ Above is a set of homegeneous equations.
$\Delta =\left|\begin{array}{ccc}1& 2& 3\\ -2& -1& 1\\ 1& 1& 2\end{array}\right|=-3+10-3=4\ne 0$Since $\Delta \ne 0,$ the system of equations has only a trivial solution i.e., $x=0,y=0,z=0$ Hence linearly independent