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Thanks for reading. I've gone through the other thread on this topic but the answer is quite different to the one I've got for the following question and I need some help in checking if my answer is correct - any help is greatly appreciated :)

The question asks to solve for "k" to make the system:

(1) Consistent with many solutions (2) Inconsistent

$$ \left[ \begin{array}{ccc|c} 1&1&k&6\\ 1&k&1&3\\ k&1&1&7 \end{array} \right] $$

I got the following row reduction:

$$ \left[ \begin{array}{ccc|c} 1&0&-1&1/(k-1)\\ 0&1&-1&-3/(k-1)\\ 0&0&k+2&6 + 2/(k-1) \end{array} \right] $$

Unfortunately I did not get the answer right. The answer is (1) none, (2) -2 and 1

My justification for this is as follows:

From the last row: if K = -2, then 0 = [something] therefore system is inconsistent when K = -2. From any row: if K = 1, then the solution will be 1/0 which doesn't exist therefore system is inconsistent when K = 1 Cannot have infinite solutions because no value of K will produce a Row of Zeroes.

I would appreciate any input on my answer/justification, especially if I've missed some vital concept.

Thank-you kindly! :)

hwt003
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1 Answers1

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The matrix row reduces to the following (without doing any 'division' by terms involving $k$:

\begin{bmatrix} 1 & 1 & k & 6\\ 0 & k-1 & 1-k & -3 \\ 0 & 0 & (k-1)(k+2) & 4-6k \end{bmatrix}

When doing row reduction, it is best not to divide by a term involving $k$ (if you divide by $k-1$ as you have above for example, then you need to consider the case $k=1$ separately).

AnyAD
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