Consider the following linear system. For what values of “K “ if any, will system has (a) no solution, (b) a unique solution ?
X1 + x2 + kx3 = 1
X1 + kx2 + x3 =1
Kx1 + x2 + x3 = -2

Question

Consider the following linear system. For what values of “K “ if any, will system has (a) no solution, (b) a unique solution ?
X1 + x2 + kx3 = 1 
X1 + kx2 + x3 =1
Kx1 + x2 + x3 = -2

ganeshie8 · Answer

$$

\left[ \begin{array}{cccc}

1 & 1 & k & 1 \
1 & k & 1 & 1 \
k & 1 & 1 & -2 \ 

\end{array} \right]

$$

ganeshie8 · Answer

familiar wid row reduction right ?

anonymous · Answer

right, but could not figure it out

ganeshie8 · Answer

lets reduce it and see wat we get

ganeshie8 · Answer

$$

\left[ \begin{array}{cccc}

1 & 1 & k & 1 \
1 & k & 1 & 1 \
k & 1 & 1 & -2 \ 

\end{array} \right]

$$ 

R2-R1
R3-kR1

$$

\left[ \begin{array}{cccc}

1 & 1 & k & 1 \
0 & k-1 & 1-k & 0 \
0 & 1-k & 1-k^2 & -2-k \ 

\end{array} \right]

$$ 

R3+R2
R2/(k-1)

$$

\left[ \begin{array}{cccc}

1 & 1 & k & 1 \
0 & 1 & -1 & 0 \
0 & 0 & 2-k-k^2 & -2-k \ 

\end{array} \right]

$$

ganeshie8 · Answer

does it look okay ?

anonymous · Answer

ok, and then ?

ganeshie8 · Answer

next is easy,
lets work infinite solutions first :-

if last row has all 0, then the system will have infinite solutions.

ganeshie8 · Answer

so, set the two elements to 0 and solve for k

$2-k-k^2 = 0$
$k = ?$


$-2-k = 0$
$k = ?$

ganeshie8 · Answer

$2-k-k^2 = 0 $
$k = -2, 1$

so this element at 33 becomes 0 when $ k = -2~~or~~ k = 1$

ganeshie8 · Answer

$-2-k = 0 $
$k = -2$

so this element at 34 becomes 0 when $ k = -2 ~~ only$

ganeshie8 · Answer

so the system will have :-

1) infinite solutions when k=-2
2) NO solutions when k = 1
3) unique solution all other cases

ganeshie8 · Answer

see if that makes some sense

anonymous · Answer

It does, tx for help

ganeshie8 · Answer

np :)

ganeshie8 · Answer

if u are familiar with rank concept, there is a nice explanation for wat we did...