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$\begingroup$

I have as far as I have shown in the picture, however I do not know how to get rid of the 3rd column above the 1. I`ve tried a couple of ways but I have only ended up messing up other columns.

enter image description here

I appreciate any help given!

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    What is the context of this picture? It kind of looks like writing a matrix in reduced-row echelon form... – 2017-02-23
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    Oh, you are finding the inverse of a matrix! Got it... How did you create the zeros in the left columns $A_1$ and $A_2$? – 2017-02-23
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    Sorry for not explaining well enough, I went onto an online calculator for inverting matrices but I do not kno how to get rid of the 1/3 and -2/3 in the 3rd column. – 2017-02-23
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    Are you familiar with matrix row operations? – 2017-02-23
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    Yes I am familiar with them, i am just not too sure how to get rid of the 3rd column with the fractions involved. – 2017-02-23

2 Answers 2

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Hint - Operate following operations.

$Row_1 - \frac 13 Row_3$

$Row_2 + \frac 23 Row_3$

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    I think you have errors with your signs. – 2017-02-23
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    @Tim Thayer no chance. If still then where? – 2017-02-23
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    My mistake. I was looking at the wrong matrix. Sorry! – 2017-02-23
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    @amar that completely went over my head, I didnt think it was that easy. Thanks for your help. – 2017-02-23
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    @Tim Thayer it's ok. – 2017-02-23
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    Mine pleasure bro. – 2017-02-23
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  • For the first row: multiply the third row by -1/3, and sum the first and third row. Then multiply the third row by -3.

  • For the second row: multiply the third row by 3/2, and sum the second and third row. Then multiply the third row by 2/3.