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Find the solution using causs Jordon elimination

$\left(\begin{array}{ccccc|c}{1} & {2} & {0} & {0} & {4} & {1} \\ {0} & {0} & {1} & {0} & {5} & {2} \\ {0} & {0} & {0} & {1} & {6} & {3} \\ {x_{1}} & {x_{2}} & {x_{3}} & {x_{4}} & {x_{5}}\end{array}\right)$

No. of free var. $=5-3=2$

Let $x_{2}=r$
$x_{5}=s$

$x_{1}+2 r+4s=1$
$x_{3}+5s=2$
$x_{4}+6s=3$

$\left[\begin{array}{l}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}} \\ {x_{5}}\end{array}\right]=\left(\begin{array}{c}{1-2 r-4 s} \\ { r} \\ {2-5 s} \\ {3-6s} \\ {s}\end{array}\right)$

$\left[\begin{array}{c}{x_{1}} \\ {x_{2}} \\ {x_{3}} \\ {x_{4}} \\ {x_{3}}\end{array}\right]=\left[\begin{array}{c}{-2} \\ {1} \\ {0} \\ {0} \\ {0}\end{array}\right]+s\left[\begin{array}{c}{-4} \\ {0} \\ {-5} \\ {-6} \\ {1}\end{array}\right]+\left[\begin{array}{c}{1} \\ {0} \\ {2} \\ {3} \\ {0}\end{array}\right]$

For what value (s) of $k,$ if any, will the systems have
(a) No solution, $(b)$ a unique solution and, $(c)$ infinitely many solutions?
$x-2 y+3 z=2$
$x+y+z=k$
$2 x-y+4 z=k^{2}$

$\left(\begin{array}{ccc|c}{1} & {-2} & {3} & {2} \\ {1} & {1} & {1} & {k} \\ {2} & {-1} & {4} & {k^{2}}\end{array}\right) \begin{array}{l}{R_{2} \rightarrow R_{2}- R_{1}} \\ {R_{3} \rightarrow R_{3} - 2 R_{1}}\end{array}$

$\left(\begin{array}{ccc|c}{1} & {-2} & {3} & {2} \\ {0} & {3} & {-2} & {k-2} \\ {0} & {3} & {-2} & {-4+k^{2}}\end{array}\right) R_{3} \rightarrow R _{3}-R_{2}$

$\left(\begin{array}{ccc|c}{1} & {-2} & {3} & {2} \\ {0} & {3} & {-2} & {k-2} \\ {0} & {0} & {0} & {k^{2}-k-2}\end{array}\right) R_{2} \rightarrow 1 / 3{R_{2}}$

$\left(\begin{array}{ccc|c}{1} & {-2} & {3} & {2} \\ {0} & {1} & {-2 / 3} & {(k-2) / 3} \\ {0} & {0} & {0} & {k^{2}-k-2}\end{array}\right)$

for unique sol.

No values for k

Inf. many sol.

$k^{2}-k-2=0$
$(k-2)(k+1)$
$k=2, k=-1$

$k \in[-1,2]$

No. sol.

$k^{2}-k - 2 \neq 0$
$(k-2)(k+1) \neq 0$
$k \neq 2, k \neq-1$

$k \in R-\{-1,2\}$