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I have an augmented matrix that represents a system of linear equations. Some components of this matrix are unknown and are represented by $ k $. My task is to find which values of $k$ would create 0, 1, or infinite solutions.

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

My strategy is to first reduce the matrix into echlon form, which will hopefully give me some insight into which values of $k$ will make the system either inconsistent (indicating no solutions), have free variables (indicating infinite solutions), or whatever is left (being unique solutions).

Step 1) Multiply Row2 and Row3 by $k$.

$$ \left[ \begin{array}{ccc|c} k&1&1&-2\\ k&k^2&k&k\\ k&k&k^2&k \end{array} \right] $$

Step 2) Subtract Row3 from Row2.

$$ \left[ \begin{array}{ccc|c} k&1&1&-2\\ 0&{k^2}-k&k-{k^2}&0\\ k&k&k^2&k \end{array} \right] $$

Step 3) Subtract Row1 from Row3

$$ \left[ \begin{array}{ccc|c} k&1&1&-2\\ 0&{k^2}-k&k-{k^2}&0\\ 0&k-1&{k^2}-1&k+2 \end{array} \right] $$

Step 4) Multiply Row3 by k

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

Step 5) Subtract Row2 from Row3

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

The matrix is now in echelon form.

If $k = 0$, then the matrix will look like this

$$ \left[ \begin{array}{ccc|c} 0&1&1&-2\\ 0&0&0&0\\ 0&0&0&0 \end{array} \right] $$

With $k=0$, we have one leading variable (Row1Col2) and one free variable (Row1Col3). This shows an infinite solution set. (A line in 3D space).

Back to the matrix in step 5. We can rewrite the equation in Row3 in general form, which looks like this: $$ ({k^3}+{k^2}-2k)z = {k^2}+2k $$ If we solve for $z$ and simplify: $$z = {1\over k-1}$$

If $k = 1$ then we will be dividing by zero, making the system inconsistent. This is clear by looking at the initial matrix, where $k$ equaling to $1$ would create 3 identical equations with 2 different constants ($1$ and $-2$).

So we have:

$k = 0$ $\to$ Infinite solutions.

$k = 1$ $\to$ Zero possible solutions

$k \neq 0$ and $k \neq 1$ $\to$ Unique solutions.

Unfortunately this isn't the answer given to me by my book. Which tells me that:

$k = -2$ $\to$ Infinite solutions.

$k = 1$ $\to$ Zero possible solutions.

$k \neq -2$ and $k \neq 1$ $\to$ Unique solutions.

Seems like I'm making a mistake finding the value of $k$ when the system has infinite solutions, getting 1 instead of -2. I'm making an attempt to use the leading/free variable techniques described in the book. I've re-read it all a few times and it seems to make sense to me.

What am I doing wrong? Thanks!

EDIT: I see that $k = -2$ would create all 0's in Row 3, which would indicate more free variables and thus infinite solutions. But why doesn't 0 work as well?

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  • $\begingroup$ In step 1, you multiplied a row by $k$. But what if $k=0$? You essentially just "deleted" a row and removed all information that it would have given you. If you decide to multiply or divide a row by an unknown number, you must treat the cases where what you are trying to multiply/divide by is zero and when it is not zero separately. $\endgroup$ – JMoravitz Dec 14 '15 at 5:13
  • $\begingroup$ Ah you are correct. Reading back it's mentioned that multiplying a row by 0 is not a valid elementary row operation. So in order to take the steps i took, i must assume that $k \neq 0$. Which leaves -2 as the option where the system has free variables and thus infinite solutions. Thanks! If you post your comment as an answer I will accept it. $\endgroup$ – David Davidson Dec 14 '15 at 5:29
  • $\begingroup$ @Variable: It's assumed for these problems that i do not know about determinants. That's for a later chapter. These problems should be solved using basic direct methods of solving systems of linear equations. $\endgroup$ – David Davidson Dec 14 '15 at 5:30
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JMoravitz gave me the answer. Step 1 is not a valid operation if $k = 0$, so if we do that operation we have to assume that $k \neq 0$, which makes my solution impossible, and only leaves -2 as the possible option in which the system has infinite solutions.

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