# How can I reduce this matrix $\left[\begin{smallmatrix}-2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\\end{smallmatrix}\right]$?

I am trying to reduce this matrix to the reduced one $$\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{bmatrix}$$

reduced to $$\begin{bmatrix} 2 & -1 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

I know how to reduce matrix. but In this case, I have no idea what happens.

I cannot get $0$'s for third row.

• Add the first and second rows to the third to start. That will get you the third row is zero.
– Arby
May 9, 2017 at 1:00
• You should show your work. It's very likely an arithmetic or otherwise simple error in the row operations, if you indeed are comfortable with row reduction. This way we can help you spot the error(s).
– Dave
May 9, 2017 at 1:02
• The matrix is singular (it’s the Laplacian of a 3-element ring graph), so it had better have at least one row of zeros when you’re done.
– amd
May 9, 2017 at 2:41

Zeros for the third row is easy, just add the first and second rows and add the result to the third row.

$\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{bmatrix}$

$R_1+R_2+R_3\to R_3\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ (-2+1+ 1) & (1+-2+1) & (1+1+-2) \\ \end{bmatrix}\implies \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}$

• I did. I add first row to second and third by multiplying 1/2 and then I got 0 -5/2 1/2 for the second raw, and 0 1/2 -5/2 for the third raw. May 9, 2017 at 1:05
• One times the first row plus one times the second row plus the third row into the third row gives zeros.
– Arby
May 9, 2017 at 1:06
• and then I cannot make third raw 0's,. May 9, 2017 at 1:06
• add second raw to third raw to make 2 - 1 -1 right ? after that I can add fist raw to thrid raw. May 9, 2017 at 1:09
• Or just do it all at once.
– Arby
May 9, 2017 at 1:10

$$\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{bmatrix}$$

Use $\rm R_3\gets R_3+R_2+R_1$

$$\begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1+1-2 & 1-2+1 & -2+1+1 \\ \end{bmatrix}$$

Then $\rm R_2\gets (2R_2+R_1)/(-3)$

$$\begin{bmatrix} -2 & 1 & 1 \\ (2-2)/(-3) & (-4+1)/(-3) & (2+1)/(-3) \\ 1+1-2 & 1-2+1 & -2+1+1 \\ \end{bmatrix}$$

Which is complete, though you can reduce $R_1$ further, via, $\rm R_1\gets (R_1+R_2)/(-2)$ if you so wish.

Pretty easy question if I understand what you're asking:

$$\begin{pmatrix} -2 & 1 & 1\\ 1 & -2 & 1\\ 1 & 1 & -2 \end{pmatrix}\Rightarrow \begin{pmatrix} -2 & 1 & 1\\ 0 & -1.5 & 1.5\\ 1 & 1 & -2 \end{pmatrix}\Rightarrow \begin{pmatrix} -2 & 1 & 1\\ 0 & -1.5 & 1.5\\ 0 & 1.5 & -1.5 \end{pmatrix}\Rightarrow \begin{pmatrix} -2 & 1 & 1\\ 0 & -1 & 1\\ 0 & 0 & 0 \end{pmatrix}$$