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I spent last hours trying to figure out how to solve the inverse matrix to this matrix: $$\begin{pmatrix} 2 &-3 & 1 \\ 1 & 2 &-1 \\ 2 & 1 & 1 \end{pmatrix}$$

The correct result should be $$\begin{pmatrix} 0.250 & 0.333 & 0.083 \\ -0.250 & 0.000 & 0.250 \\ -0.250 & -0.667 & 0.583 \end{pmatrix}$$

However, I am still unable to get there. Here how I tried it (with using the Gaussian Elimination Rule):

$$\begin{multline} \left( \begin{array}{ccc|ccc} 2 & -3 & 1 & 1 & 0 & 0 \\ 1 & 2 & -1 & 0 & 1 & 0 \\ 2 & 1 & 1 & 0 & 0 & 1 \end{array} \right) \overset{[1] - 2[2] \rightarrow [2]}{\Longrightarrow} \left( \begin{array}{ccc|ccc} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 2 & 1 & -2 & 0 \\ 2 & 1 & 1 & 0 & 0 & 1 \end{array} \right) \overset{[1] - [3] \rightarrow [3]}{\Longrightarrow} \\ \left( \begin{array}{ccc|ccc} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 2 & 1 & -2 & 0 \\ 0 & -4 & 0 & 1 & 0 & -1 \end{array} \right) \overset{4[2] - 7[3] \rightarrow [3]}{\Longrightarrow} \left( \begin{array}{ccc|ccc} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 2 & 1 & -2 & 0 \\ 0 & 0 & 8 & -3 & -8 & 7 \end{array} \right) \overset{4[2] - [3] \rightarrow [2]}{\Longrightarrow} \\ \left( \begin{array}{ccc|ccc} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & 8 & -3 & -8 & 7 \end{array} \right) \overset{8[1] - 3[2] \rightarrow [1]}{\Longrightarrow} \left( \begin{array}{ccc|ccc} 16 & -24 & 0 & 11 & 8 & -7 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & 8 & -3 & -8 & 7 \end{array} \right) \Longrightarrow \\ \left( \begin{array}{ccc|ccc} 2 & -3 & 0 & \tfrac{11}{8} & 1 & \tfrac{-7}{8} \\ 0 & -7 & 0 & \tfrac{7}{4} & 0 & \tfrac{-7}{4} \\ 0 & 0 & 8 & -3 & -8 & 7 \end{array} \right) \overset{7[1] - 3[2] \rightarrow [1]}{\Longrightarrow} \left( \begin{array}{ccc|ccc} 14 & 0 & 0 & \tfrac{35}{8} & 7 & \tfrac{-7}{8} \\ 0 & -7 & 0 & \tfrac{7}{4} & 0 & \tfrac{-7}{4} \\ 0 & 0 & 8 & -3 & -8 & 7 \end{array} \right) \Longrightarrow \\ \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0.3125 & 0.5 & -0.0625 \\ 0 & 1 & 0 & -0.25 & 0 & 0.25 \\ 0 & 0 & 1 & -0.375 & -1 & 0.875 \end{array} \right) \end{multline}$$

(Original images: one, two)

I would be very grateful guys for helping me to figure out what I am doing wrong, there's always something why the whole inverse matrix is not correct.

Thank you very much

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  • $\begingroup$ alright I've done this on scratch paper and I've revealed the row operations on a few comments below someone's answer. However, I think the number on the top left is wrong... the answer to the entire first row should be $\frac{1}{4}. \frac{1}{3}$ and $\frac{2}{3}$ the answer to the second and third rows are correct $\endgroup$
    – usukidoll
    May 1, 2014 at 10:14
  • $\begingroup$ I have attempted to improve the readability of your question by introducing MathJax. Please check to ensure that I have not made any errors in transcribing your images. $\endgroup$
    – user642796
    May 2, 2014 at 9:09

2 Answers 2

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At first step [1]-2[2], i think the entry (2,3) is 3, while you wrote 2.

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  • $\begingroup$ No it's "the first row" - "2x second row" and save it (write it) to the second row. $\endgroup$
    – user984621
    May 1, 2014 at 9:31
  • $\begingroup$ are you trying to use Gauss elimination to obtain the row echelon form? Do you know what an upper triangular matrix is? That's what a row echelon form matrix looks like with the exception that the top left must always be a 1. $\endgroup$
    – usukidoll
    May 1, 2014 at 9:33
  • $\begingroup$ try do these row operations $-r_1+r_3 \rightarrow r_3$........ $-r_1+2r_2 \rightarrow r_2$... sorry I'm doing this in my head.. $\endgroup$
    – usukidoll
    May 1, 2014 at 9:35
  • $\begingroup$ Yes, I am doing it through the Gauss elimination. $\endgroup$
    – user984621
    May 1, 2014 at 9:37
  • $\begingroup$ ok... I'm doing this on paint right now ... try these row operations for now $ -r_1+ r_3 \rightarrow r_3$.........$-r_1+2r_2 \rightarrow r_2$...........$r_2 \leftrightarrow r_3$..........$-7r_2+4r_3 \rightarrow r_3$....$\frac{1}{4}r_2 \rightarrow r_2$......$\frac{-1}{12} r_3 \rightarrow r_3$ wait a sec I'm missing something.. umm can you tell me what the right side of the matrix is? I think row operations must also happen as well. so we are doing row operations for those two matrices... appears to be an augmented matrix. A l B $\endgroup$
    – usukidoll
    May 1, 2014 at 9:39
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It's very hard to avoid unimportant bugs in this process. Here's a corrected version. (I'm not keen on performing more than one row operaion at once.) The first error is what BTTD pointed out (purely arithmetic) and it leads to other problems down the track.

\begin{align*} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 1 & 2 & -1 & 0 & 1 & 0 \\ 2 & 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix} &\xrightarrow{R_2 \gets -2 R_2} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ -2 & -4 & 2 & 0 & -2 & 0 \\ 2 & 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix} \\ &\xrightarrow{R_2 \gets R_2+R_1} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 3 & 1 & -2 & 0 \\ 2 & 1 & 1 & 0 & 0 & 1 \\ \end{bmatrix} \\ &\xrightarrow{R_3 \gets -R_3} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 3 & 1 & -2 & 0 \\ -2 & -1 & -1 & 0 & 0 & -1 \\ \end{bmatrix} \\ &\xrightarrow{R_3 \gets R_3+R_1} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 3 & 1 & -2 & 0 \\ 0 & -4 & 0 & 1 & 0 & -1 \\ \end{bmatrix} \\ &\xrightarrow{R_3 \gets 7 R_3} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 3 & 1 & -2 & 0 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_3 \gets R_3-4R_2} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -7 & 3 & 1 & -2 & 0 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_2 \gets 4 R_2} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -28 & 12 & 4 & -8 & 0 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_2 \gets R_2+R_3} \begin{bmatrix} 2 & -3 & 1 & 1 & 0 & 0 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_1 \gets 12 R_1} \begin{bmatrix} 24 & -36 & 12 & 12 & 0 & 0 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_1 \gets R_1+R_3} \begin{bmatrix} 24 & -36 & 0 & 15 & 8 & -7 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_1 \gets 7 R_1} \begin{bmatrix} 168 & -252 & 0 & 105 & 56 & -49 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_1 \gets R_1-9R_2} \begin{bmatrix} 168 & 0 & 0 & 42 & 56 & 14 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_1 \gets \tfrac{1}{168} R_1} \begin{bmatrix} 1 & 0 & 0 & 1/4 & 1/3 & 1/12 \\ 0 & -28 & 0 & 7 & 0 & -7 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_2 \gets -\tfrac{1}{28} R_2} \begin{bmatrix} 1 & 0 & 0 & 1/4 & 1/3 & 1/12 \\ 0 & 1 & 0 & -1/4 & 0 & 1/4 \\ 0 & 0 & -12 & 3 & 8 & -7 \\ \end{bmatrix} \\ &\xrightarrow{R_3 \gets -\tfrac{1}{12} R_3} \begin{bmatrix} 1 & 0 & 0 & 1/4 & 1/3 & 1/12 \\ 0 & 1 & 0 & -1/4 & 0 & 1/4 \\ 0 & 0 & 1 & -1/4 & -2/3 & 7/12 \\ \end{bmatrix} \\ \end{align*}

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