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I have the following matrix and have to see if it has solutions depending on $a$. My solution:

$M= \left[ {\begin{array}{cc} a & a^2 &| &1 \\ -1 & -1& | & -a \\ 1 & a & | & a \end{array} } \right] $

My attemp was: Changing first with third line

$= \left[ {\begin{array}{cc} 1 & a & | & a \\ -1 & -1& | & -a \\ a & a^2 &| &1 \\ \end{array} } \right] $

Add the first row to the second one

$= \left[ {\begin{array}{cc} 1 & a & | & a \\ 0 & a-1& | & 0 \\ a & a^2 &| &1 \\ \end{array} } \right] $

Add the $-a$ of the first line to the third line

$= \left[ {\begin{array}{cc} 1 & a & | & a \\ 0 & a-1& | & 0 \\ 0 & 0 &| &1-a^2 \\ \end{array} } \right] $

From $0=1-a^2$there we can obtain that the LES has a solution if $a=+/-1$

Is this a valid solution to the problem and is there a free variable?

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If $a=\pm 1$, then indeed we have a solution.

Now $a=1$, the second row become the zero row, and hence there is a free variable. Can you write down the general solution?

If $a=-1$, the second row is not a zero row, there is no free variable. It has a unique solution.

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  • $\begingroup$ but for $a=-1$ $x_2$ has to be $0$ so the row becomes a zero row again, doesn't it? $\endgroup$ – Segonut Jan 10 at 13:48
  • $\begingroup$ $x_2=0$ is true, but the second row is $[0, -2, 0]$ which is not a zero row. $\endgroup$ – Siong Thye Goh Jan 10 at 13:50
  • $\begingroup$ okay, thank you! And the transformations are correct? $\endgroup$ – Segonut Jan 10 at 13:50
  • $\begingroup$ And for $a=1$ the general solution should be $L=[1-t, t]$, shouldn't it? Because $x_1=1-x_2$ and $x_2=t$ (or any other name)? And for $a-1$ the solution is $[a,0]$, right? $\endgroup$ – Segonut Jan 10 at 13:54
  • $\begingroup$ Your Gaussian elimination step looks fine. if $a=1$, the general solution is indeed $(1-t,t)$ where $t$ is a free parameter. If $a=-1$, we have $x_2=0$, and the solution is $(a,0)=(-1,0)$. $\endgroup$ – Siong Thye Goh Jan 10 at 14:03
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\begin{align} M= \left[\begin{array}{cc|c} a & a^2 & 1 \\ -1 & -1 & -a \\ 1 & a & a \end{array}\right] &\to \left[\begin{array}{cc|c} 1 & a & a \\ -1 & -1 & -a \\ a & a^2 & 1 \end{array}\right] &&R_1\leftrightarrow R_3 \\[4px]&\to \left[\begin{array}{cc|c} 1 & a & a \\ 0 & a-1 & 0 \\ 0 & 0 & 1-a^2 \end{array}\right] &&\begin{aligned}R_2&\gets R_2+R_1\\R_3&\gets R_3-aR_1\end{aligned} \end{align} Your work was pretty good.

Now you have to distinguish the cases $a=1$, $a=-1$, $a\ne\pm1$

Case $a=1$

The matrix becomes $$ \left[\begin{array}{cc|c} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ and the system has infinitely many solutions $$ \begin{bmatrix} 1-h \\ h \end{bmatrix}= \begin{bmatrix} 1 \\ 0 \end{bmatrix} + h\begin{bmatrix} -1 \\ 1 \end{bmatrix} $$

Case $a=-1$

The matrix becomes $$ \left[\begin{array}{cc|c} 1 & -1 & -1 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right] \to \left[\begin{array}{cc|c} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right] $$ and the system has a single solution $\begin{bmatrix} -1 \\ 0 \end{bmatrix}$.

Case $a\ne\pm1$

No solution.

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