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$$\begin{cases} 2x_1-x_2= dx_1 \\ 2x_1-x_2+x_3=dx_2 \\ -2x_1+2x_2+x_3=dx_3 \end{cases} $$

a) Is it possible for the system to be inconsistent? Explain?

b) For what values of d will the system have infinitely many solutions?

c) Solve the system when it has infinitely many solutions?

For my solution in part a), $$ \left[ \begin{array}{ccc|c} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{array} \right] $$

Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?

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  • $\begingroup$ Welcome to Math SE! Links to images are not accessible to screen readers. You can instead use MathJax to type mathematics on this site – in particular the 'Systems of equations' part. $\endgroup$ – Toby Mak Sep 1 '18 at 4:22
  • $\begingroup$ For my solution in part a) The rest of the question appears to center on this part, which you haven't posted. Please see How do I ask a good question? and add at least some minimal context. $\endgroup$ – dxiv Sep 1 '18 at 4:28
  • $\begingroup$ Hi, I have edited my question accordingly, thank you for the suggestion! $\endgroup$ – Cheryl Sep 1 '18 at 4:58
  • $\begingroup$ You should be able to tell that the system can’t be inconsistent by inspection: it’s homogeneous, so $x_1=x_2=x_3=0$ is always a solution. $\endgroup$ – amd Sep 1 '18 at 8:59
  • $\begingroup$ How were you able to fully reduce the coefficient matrix without knowing what $d$ is? You might have divided by zero somewhere along the way, which would make the reduction invalid. $\endgroup$ – amd Sep 1 '18 at 9:00
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Using the Gaussian elimination method to solve a set of linear equations,

From the equations, you have given,

\begin{cases} 2x_1-x_2= dx_1 \\ 2x_1-x_2+x_3=dx_2 \\ -2x_1+2x_2+x_3=dx_3 \end{cases}

We can arrive at this augmented matrix,

\begin{bmatrix} \begin{array}{ccc|c} -d+2&-1&0&0\\ 2&-1-d&1&0\\ -2&2&1-d&0 \end{array} \end{bmatrix}

Using row transformations,

\begin{bmatrix} \begin{array}{ccc|c} -2 & 2 & 1-d &0 \\ 2 & -1-d & 1 &0\\ -d+2 & -1 & 0 &0\\ \end{array} \end{bmatrix}

\begin{equation} \downarrow \end{equation}

\begin{bmatrix} \begin{array}{ccc|c} -2 & 2 & 1-d &0 \\ 0 & 1-d & 2-d &0\\ 0 & 0 & (1-d)(2-d) -2(2-d) &0\\ \end{array} \end{bmatrix}

For it to have infinite solutions,

\begin{equation} (1-d)(2-d) -2(2-d) = 0 \end{equation}

\begin{equation} d = 2, -1 \end{equation}

If $d$ takes the above value, then you will end up with a free variable ($x_3$)

\begin{equation} x_2 = \frac{(d-2)x_3}{1-d} \end{equation}

\begin{equation} x_1 = \frac{1}{2}(2x_2 + (1-d)x_3) \end{equation}

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  • $\begingroup$ May I know how you derived your last row to that? I can't seem to eliminate my first 2 columns of the last row to = 0... $\endgroup$ – Cheryl Sep 1 '18 at 9:39
  • $\begingroup$ Use $R_2 = R_1 + R_2$ ,$R_3 = (-d+2)R_1 + 2R_3$, then $R_3 = R_3 - 2R_2$ then it reduces to this form. $\endgroup$ – Lokesh Kumar Sep 1 '18 at 9:49
  • $\begingroup$ Got it, thank you for your help! $\endgroup$ – Cheryl Sep 1 '18 at 10:00

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