I am trying to solve for the inverse for a given $3x3$ matrix. The matrix looks like this: \begin{pmatrix}1&0&1\\ \:a&0&1\\ \:1&a&0\end{pmatrix} I have tried solving it through carrying out row operations on the matrix above whilest in parallel carrying out the same operations on a $3x3$ Identity matrix. I am worried that I might be preforming "undefined" or "non-approved" row operations on $A$ and that it, is the cause of the incorrect result. What i got so far looks like this:

\begin{pmatrix}1&0&1\\ \:\:\:\:\:a&0&1\\ \:\:\:\:\:1&a&0\end{pmatrix} $R_2\to R_2 -aR_1$

$R_3\to R_3-R_1$ \begin{pmatrix}1&0&1\\ \:\:\:\:\:\:0&0&1-a\\ \:\:\:\:\:\:0&a&-1\end{pmatrix}

Assuming $a\neq 1$

Switch $R_2\leftrightarrow R_3$

\begin{pmatrix}1&0&1\\ \:\:\:\:\:0&a&-1\\ \:\:\:\:\:0&0&1-a\end{pmatrix}

$R_3\to \frac{1}{1-a}R_3$\begin{pmatrix}1&0&1\\ \:\:\:\:\:\:0&a&-1\\ \:\:\:\:\:\:0&0&1\end{pmatrix}

$R_2\to R_2+R_3$

$R_1\to R_1-R_3$

\begin{pmatrix}1&0&0\\ \:\:\:\:\:\:\:0&a&0\\ \:\:\:\:\:\:\:0&0&1\end{pmatrix} $R_2\to R_2\cdot(1/a)$ \begin{pmatrix}1&0&0\\ \:\:\:\:\:\:\:0&1&0\\ \:\:\:\:\:\:\:0&0&1\end{pmatrix}

This it what i am getting, if i do the same elementry row operations in the same sequence to the Identity matrix i get an incorrect result. What am i doing wrong? (Sorry about the messy notation, first time poster) Kind regards and many thanks!


$a≠0,1$ or the matrix will be singular. The row transformations look okay. The inverse comes out to be $\begin{pmatrix}\frac1{1-a}&\frac1{a-1}&0\\ \:\frac1{a(a-1)}&\frac1{a(1-a)}&\frac1a\\ \:\frac a{a-1}&\frac1{1-a}&0\end{pmatrix}$.


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