Answer
(i) Subtracting the second equation from the first equation and adding the third to the difference results in the equation $1\neq 0$
(ii) 0 from the last equation on the RHS must be replaced with $-1$ to get a non-singular system of linear equations.
(iii) One of the solutions is $u=2$,$v=1$, $w=-1$
Work Step by Step
(i) Given a system of linear equations is-
$u +1v + 1w = 2$
$u +2v+3w =1$
$\hspace{0.8cm}1v+2w=0$
Consider the LHS -
Subtracting the second from the first equation we get
$-1v-2v$
Adding the LHS of the third equation gives $0$
Applying the same operations to the RHS, we get $1$, Hence LHS$\neq$RHS, or $1\neq0$
(ii) To make the RHS $0$ we need the last $0$ to be $-1$
(iii) Using the column picture, we must find a combination of the three coefficient columns on the LHS which results in the RHS, hence plugging in $u=2$, $v=1$, and $w=-1$ should make this possible.
