For example,
$3=3I$
$=3\begin{bmatrix} 1 & 0 \\ 0 &1 \\ \end{bmatrix}$
$=\begin{bmatrix} 3 & 0 \\ 0 &3 \\ \end{bmatrix}$
and $\begin{vmatrix} 3 & 0 \\ 0 & 3 \\ \end{vmatrix}=9$
Does this mean $3=9$ ? Am I confusing determinant with "value of matrix"?
On
If $A$ is a $n\times n$-matrix and $t$ a real number, we have
$$\det(tA)=t^n\cdot A$$
So, in fact, you confused the determinant with a multiplication by a scalar.
On
For any scalar $\;k\;$ and any $\;n\times n\;$ matrix $\;A\;$ , we have that
$$|kA|=\det (kA)=k^n\det A=k^n|A|$$
so in the example you wrote in your question:
$$|3I_2|=3^2|I_2|=9\cdot1=9$$
and yes: you were confusing things.
On
Replacing 1 by I is a delicate thing.
$I$, the $n\times n$ identity matrix, enjoys many properties similar to the unit element in ${\mathbb R}$. If $A$ is any $n\times n$ matrix then $AI=IA=A$ and if $p(t)=p_0+ p_1 t + \cdots + p_d t^d$ is a (usual) polynomial, it makes sense to consider $p(A)= p_0 I + p_1 A + \cdots + p_d A^d$ as a matrix valued polynomial. If, e.g. $p(t)=\det(A-t I)$ is the characteristic polynomial of $A$ then the (non-trivial) Hamilton-Cayley theorem tells us that $p(A)=0$ (the zero matrix on the RHS).
But alas, 1 and I live in different spaces so they are not equal, so your calculation does not make sense.
On
No it is usually not correct to replace $3$ by $3I$. The former is the number $3$, while the latter is a matrix of some (possibly unspecified) dimension with $3$ on the diagonal. This is not the same object (except possibly if the dimension happens to be one, but this is a bit of a degenerate case).
There is some kind of an 'exception' to this though, which may be the cause of your questions:
Given a polynomial, say $p(x)= x^2 + 2x + 3$, one has that for a matrix $A$, $$p(A)= A^2 + 2A + 3 I.$$ Thus here it seems that $3$ is replaced by $3I$. Yet, the confusion should go away when one recalls that $p(x)= x^2 + 2x + 3$ is short for $p(x)= x^2 + 2x^1 + 3x^0$, and $A^0 = I$.
On
The rule in the case of a constant (k) in one row is
$\left|\begin{matrix} a_{11} & a_{12}&\ldots&a_{1n} \\ \vdots &\vdots&\vdots&\vdots \\ \color{red}{k}\cdot a_{s1} & \color{red}{k}\cdot a_{s2}&\ldots& \color{red}{k}\cdot a_{sn}\\ \vdots &\vdots&\vdots&\vdots \\ a_{m1} & a_{m2}&\ldots&a_{mn} \end{matrix}\right|=\color{red}{k}\cdot \left|\begin{matrix} a_{11} & a_{12}&\ldots&a_{1n} \\ \vdots &\vdots&\vdots&\vdots \\ a_{s1} & a_{s2}&\ldots& a_{sn}\\ \vdots &\vdots&\vdots&\vdots \\ a_{m1} & a_{m2}&\ldots&a_{mn} \end{matrix}\right|$
Or similarly: The rule in the case of a constant (k) in one column is
$\left|\begin{matrix} a_{11} & \color{red}{k}\cdot a_{12}&\ldots&a_{1n} \\ \vdots &\vdots&\vdots&\vdots \\ a_{s1} & \color{red}{k}\cdot a_{s2}&\ldots& a_{sn}\\ \vdots &\vdots&\vdots&\vdots \\ a_{m1} & \color{red}{k}\cdot a_{m2}&\ldots&a_{mn} \end{matrix}\right|=\color{red}{k}\cdot \left|\begin{matrix} a_{11} & a_{12}&\ldots&a_{1n} \\ \vdots &\vdots&\vdots&\vdots \\ a_{s1} & a_{s2}&\ldots& a_{sn}\\ \vdots &\vdots&\vdots&\vdots \\ a_{m1} & a_{m2}&\ldots&a_{mn} \end{matrix}\right|$
If you have $p$ rows/columns (distinct cases) with a costant ($k$) then the determinant has to be multiplied by k raised to power $p$.
You have $p=2$ rows (or columns) with the factor $k=3$. Therefore
$|A|=3^2\cdot |I|$
If you know already that you're working with linear transformations of a particular vector space (which form an associative algebra over the scalar field), then it is customary to write simply $3$ for the transformation that multiplies every vector by $3$, which is the same as $3I$.
So if that is your context, then $3$ and $3I$ is the same thing. That doesn't mean that the matrix $3I$ is the same as the number three, just that the symbol "$3$" is sometimes used to denote that matrix in addition to the number.
But the real problem in your calculation is where you apparently conclude from $|3I|=9$ that $3I=9$.
A matrix is not equal to its determinant -- you're getting those determinant bars out of nowhere, and you have no right to assume that $A=|A|$ even in a context where you use bare numbers as shorthand for certain matrices.
If it's given that you're working in the algebra of linear transformations of $\mathbb R^2$, then the best the various abbreviations will let you conclude is $$ \det(3) = 9 $$ which looks strange but is not the kind of contradiction that $3=9$ is.