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Suppose that $\det(A) = 1$ , show $A=I$?

How to proof this question??

I tried like that

$\det(A)=1 , \det(I)=1 $

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    $\begingroup$ This is false, put whatever you want above the diagonal of an upper triangular matrix with 1's on the diagonal $\endgroup$ – qbert Oct 23 '17 at 16:29
  • $\begingroup$ Even if you constrain $A$ to be diagonal, this is false. Consider e.g. the $2 \times 2$ matrix $A = -I$. More generally, let $A$ be any diagonal matrix, the product of whose diagonal entries equals $1$. $\endgroup$ – Bungo Oct 23 '17 at 16:30
  • $\begingroup$ $$ \det \left[ \begin{array}{cc} 7 & 8 \\ 6 & 7 \end{array} \right] = 1. $$ There are many matrices whose determinant is $1. \qquad$ $\endgroup$ – Michael Hardy Oct 23 '17 at 16:34
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    $\begingroup$ It is more or less like saying that if a number is odd, it is $37$. $\endgroup$ – Jack D'Aurizio Oct 23 '17 at 16:35
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    $\begingroup$ $A=I\implies \det(A)=1$, not $A=I\iff \det(A)=1$ $\endgroup$ – MCCCS Oct 23 '17 at 16:35
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It's not true.

Just consider the matrix

$$A=\begin{bmatrix}1&0\\1&1\end{bmatrix}$$

Clearly $\det(A)=1$, but $A\neq I$

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Counterexample $$A=\begin{pmatrix}2 & 1 \\ 3 & 2\end{pmatrix}$$

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