# Matrices how to prove $A^{-1} = A$

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?

$$A^{-1}$$

$$x_1$$ = $$2/49$$

$$A$$

$$x_1$$ = $$2/7$$

• if you want to prove $A=A^{-1}$, did you try computing $A\times A$? – J. W. Tanner May 26 at 15:15
• You probably forgot the 1/7 in front of the matrix. – N. S. May 26 at 15:21
• 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 $A$. – amd May 27 at 0:53

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.