1
$\begingroup$

I have to find a row-reduced matrix which is row-equivalent to

$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right]$$

On one hand

$$\left[\begin{array}{ccc} i & -(1+i) & 0 \\ 1 & -2 & 1 \\ 1 & 2i & -1\\ \end{array}\right] \\ \xrightarrow{R_1 \leftrightarrow R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ i & -(1+i) & 0 \\ 1 & 2i & -1\\ \end{array}\right]$$

$$\xrightarrow{R_2 \gets -iR_1 + R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 1 & 2i & -1\\ \end{array}\right] $$

$$\xrightarrow{R_3 \gets -R_1 + R_3} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & -1+i & -i \\ 0 & 2+2i & -2\\ \end{array}\right] $$

$$\xrightarrow{\ \frac{1}{(-1+i)}R_2} \left[\begin{array}{ccc} 1 & -2 & 1 \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$

$$\xrightarrow{\ R_1\gets 2R_2+R_1} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 2+2i & -2\\ \end{array}\right] $$

$$\xrightarrow{\ R_3\gets -(2+2i)R_2+R_3} \left[\begin{array}{ccc} 1 & 0 & i \\ 0 & 1 & 1/(-1-i) \\ 0 & 0 & -4\\ \end{array}\right] $$

and this last matrix is equivalent to the identity matrix $I_{3x3}$.

On the other hand I reduced it with other row operations and concluded the same as WOLFRAM ALPHA. Which one is wrong? Any help is appreciated.

$\endgroup$
  • $\begingroup$ I am not getting what you want to describe $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 16:38
  • $\begingroup$ You want to convert it into identity matrix of 3×3? $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 16:38
  • $\begingroup$ @KanwaljitSingh I end up with two different row reduced matrices, one being the identity matrix and the other one shown in the link to wolfram alpha. This means one of them must be wrong right? $\endgroup$ – AndreGSalazar Apr 18 '17 at 16:42
  • $\begingroup$ Looks like a simple arithmetic error when you computed the (3,3) element in the last step. You probably forgot to include a minus sign. $\endgroup$ – amd Apr 18 '17 at 16:55
  • $\begingroup$ @amd indeed $\frac{-(2+2i)}{-(1+i)}=2$ hence $(3,3)=0$, thank you $\endgroup$ – AndreGSalazar Apr 18 '17 at 17:05
0
$\begingroup$

Your mistake is in 4th step. When you are taking $(-1+i)$ common from $R_2$. Then element at 2nd row and 3rd column is wrong I think.

$\endgroup$
  • $\begingroup$ Sorry, I meant to multiple by $\frac{1}{-1+i}$. Already edited. $\endgroup$ – AndreGSalazar Apr 18 '17 at 16:58
  • $\begingroup$ But its still wrong. $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 17:24
  • $\begingroup$ Then it becomes $\frac{i}{1-i}$. $\endgroup$ – Kanwaljit Singh Apr 18 '17 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.