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For what value of k does the system of equations not have a unique solution?

$$ \left\{ \begin{array}{c} x-2y+2z=0 \\ 2x+ky-z=3 \\ x-y+3z=-5 \end{array} \right. $$

I know that this means I have to find the value(s) of k where the system of equations has either no solutions or an infinite number of them.

I converted the above system into a matrix and tried to simplify it into Row Reduced Echelon Form (rref), however, had no luck because of the one variable.

Here's what I did.

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\2&k&-1&3\\1&-1&3&-5\end{array}\right]$$

$$R_1 - R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\2&k&-1&3\\0&-1&-1&5\end{array}\right]$$

$$2*R_1$$

$$=\left[\begin{array}{ccc|c}2&-4&4&0\\2&k&-1&3\\0&-1&-1&5\end{array}\right]$$

$$R_1 - R_2$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-4-k&5&-3\\0&-1&-1&5\end{array}\right]$$

$$(-1)*R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-4-k&5&-3\\0&1&1&-5\end{array}\right]$$

$$4*R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-4-k&5&-3\\0&4&4&-20\end{array}\right]$$

$$R_2 + R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-k&9&-23\\0&4&4&-20\end{array}\right]$$

$$4*R_2$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-4k&36&-92\\0&4&4&-20\end{array}\right]$$

$$k*R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&-4k&36&-92\\0&4k&4k&-20k\end{array}\right]$$

$$R_2 + R_3$$

$$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&0&4k+36&-92-20k\\0&4k&4k&-20k\end{array}\right]$$

I've tried numerous other methods relying entirely on guess and check, but have not had any success. I want to know whether I am on the right track, and if so how I should continue to get to the final answer. I would also like to know whether or not there is a more systematic way of simplifying matrices into RREF form.

Any help will be greatly appreciated, thanks in advance.

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  • $\begingroup$ Keep going: do $4R_2+kR_3$ $\endgroup$ – Santiago Canez Aug 14 '15 at 2:27
  • $\begingroup$ Now what? I'm sorry if it's a silly question - am a tenth grader who's trying to self-study :P $\endgroup$ – StopReadingThisUsername Aug 14 '15 at 2:33
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    $\begingroup$ It would have been better to put the result into the third row rather than the second as you did. Or, you can now simply switch the second and third rows. From the resulting form you should be able to determining when there will be a unique solution. You can read about "Gaussian elimination" to learn more about this type of procedure. $\endgroup$ – Santiago Canez Aug 14 '15 at 2:36
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"For what value of $k$ does the system not have a unique solution"

This would include both the case of infinitely many solutions and also the case of no solution.

A much easier approach than performing row operations is the following:

Let us express this as a matrix equation $Ax=b$ instead of as a system of equations. There is a theorem:

$Ax=b$ has a unique solution if and only if $A$ is invertible.

There is another theorem:

$A$ is invertible if and only if $\det(A)\neq 0$

So, the value(s) for $k$ such that the system does not have a unique solution are those such that $\det(A)=0$.

$\det(A)=1(k\cdot 3 - (-1)\cdot(-1)) - (-2)(2\cdot 3 - (-1)\cdot 1) + 2(2\cdot (-1) - k\cdot 1)$

$=(3k-1)+(14)+(-4-2k)$

$= k+9$

So, if $\det(A)=0$, you would have $k+9=0$, which implies that $k=-9$


As for your method and current work, you are nearly there. You left off at:

$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&0&4k+36&-92-20k\\0&4k&4k&-20k\end{array}\right]$

Now, rowswap to get $R_2\leftrightarrow R_3$

$=\left[\begin{array}{ccc|c}1&-2&2&0\\0&4k&4k&-20k\\0&0&4k+36&-92-20k\end{array}\right]$

It will be that there are no solutions when the final row looks like $[0~0~0~|~n]$ with $n\neq 0$, or infinitely many solutions if the final row looks like $[0~0~0~|~0]$ (assuming that there are no other $[0~0~0~|~n]$ rows elsewhere after row reduction is complete)

To get the third entry of the last row equal to zero, that would happen when $4k+36=0$ which occurs when $k=-9$.

There is a slight error in your work which is difficult to notice, and that is the step where you multiplied a row by $k$. In the case that you allow $k=0$, you have in effect destroyed the information that that row could have given you. Remember that you are only allowed to multiply rows by nonzero numbers. When multiplying by unknown quantities, you should be careful after reaching a final answer that doing so didn't cause a problem (like it did this time). Your work would imply that by setting $k=0$ the second row becomes all zeroes, implying infinitely many solutions, but that is not the case (as shown by my first method).

Instead, of doing $4R_2\mapsto R_2$, followed by $kR_3\mapsto R_3$, followed by $R_2+R_3\mapsto R_2$., you would have avoided this error by simply doing $4R_2+kR_3\mapsto R_3$. Doing so wouldn't have caused any loss of information.

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  • $\begingroup$ After the step $4*R_2$, make two cases. Case-1 should be as you and @JMoravitz talked about (considering $k \neq 0$). Other case, consider $k=0$ and proceed further. $\endgroup$ – Rajat Aug 14 '15 at 3:50

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