Question
If $\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}6 \\ 3 \\ 2\end{array}\right]$, then the value of $(2 x+y-z)$ is
✓
Answer
$\left[\begin{array}{lll}1 & 1 & 1 \\0 & 1 & 1 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\y \\z\end{array}\right]=\left[\begin{array}{l}6 \\3 \\2 \end{array}\right] $
$\therefore x+y+z=6 .....(i)$
$y+z=3........(ii)$
$ z=2.......(iii)$
$\Rightarrow y+2=3$
$[$Using $(ii)$ and $(iii)]$
$\Rightarrow y=1$
$\Rightarrow x+1+2=6$
$\Rightarrow x=3$
$[$Using $(i), (iii)$ and $(iv)]$
So, $2 x+y-z$
$=(2 \times 3)+1-2$
$=6+1-2$
$=5$
$\therefore x+y+z=6 .....(i)$
$y+z=3........(ii)$
$ z=2.......(iii)$
$\Rightarrow y+2=3$
$[$Using $(ii)$ and $(iii)]$
$\Rightarrow y=1$
$\Rightarrow x+1+2=6$
$\Rightarrow x=3$
$[$Using $(i), (iii)$ and $(iv)]$
So, $2 x+y-z$
$=(2 \times 3)+1-2$
$=6+1-2$
$=5$
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