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Apologies mix up from earlier the wrong values where placed in $x_2$ and $x_3$.

Question 1

Proof that the following is true for matrix $A$, $A^{-1}$ = $A^{T}$ = $A$

$A$= $$ 1/7 \begin{pmatrix} 2 & 3 & 6 \\ 3 & -6 & 2 \\ 6 & 2 & -3 \\ \end{pmatrix} $$

$A^T$= $$ 1/7 \begin{pmatrix} 2 & 3 & 6 \\ 3 & -6 & 2 \\ 6 & 2 & -3 \\ \end{pmatrix} $$ The determinant is $343$

The rule has already been applied to the matrix $(+ - +) $

$A^{-1}$=

$$ 1/343 \begin{pmatrix} 14 & 21 & 42 \\ -14 & -21 & -42 \\ 42 & 14 & 21 \\ \end{pmatrix} $$

This is as far as I can go the identity rule is not producing $1$ in the diagonal how can I solve it from here?

$A^{-1}$

$x_1$ = $2/49$

$A$

$x_1$ = $2/7$

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  • $\begingroup$ if you want to prove $A=A^{-1}$, did you try computing $A\times A$? $\endgroup$ – J. W. Tanner May 26 at 15:15
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    $\begingroup$ You probably forgot the 1/7 in front of the matrix. $\endgroup$ – N. S. May 26 at 15:21
  • $\begingroup$ If $\det A = 343$, then you can’t possibly have $AA^T=I$ since $\det(AA^T)=\det(A)^2$. You’ve neglected to account for the common denominator that you pulled out of $A$. $\endgroup$ – amd May 27 at 0:53
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To check that $A^{-1}=A$, you don't need to "calculate" $A^{-1}$. If $A^{-1}=A$, then $A^2=A^{-1}A=I$; and viceversa, if $A^2=I$, then you know that $A^{-1}=A$. Here you can calculate directly that $A^2=I$.

Now, in light of the above, your calculation of $A^{-1}$ is wrong. You don't say what computations you made, so I cannot comment on that.

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