When does trace and determinant of a 2 x 2 matrix equal each other? (Linear Algebra) 
*

*Background Information: 


I am new to linear algebra, and I recently came across this homework question that I am confused about. I appreciate any explanation that can help me improve my solution.


*

*Question: 


What condition on the entries of a 2x2 matrix A means Tr(A) = det(A)? Provide
two distinct examples of 2x2 matrices which satisfy this.


*

*My approach (Not Complete):


Considering the following 2 x 2 matrix, the det(A) = 4, and Tr(A) = 4 
\begin{bmatrix}
    2       & 0\\
    0    &   2
\end{bmatrix}
However, considering this 2 x 2 matrix, the det(A) = 9, and Tr(A) = 6 
\begin{bmatrix}
    3       & 0\\
    0    &   3
\end{bmatrix}
I think the condition would be having 2 x 2 matrix such that the matrix is 
(symmetric) and (n = 2).
\begin{bmatrix}
    n       & 0\\
    0    &   n
\end{bmatrix}
My solution makes sense, but I feel it is incomplete. Am I missing a key point or a concept that I can add to my answer? 



*

*Edited:


I have tried this solution with so many numbers and it seems to work. Would this be an acceptable solution?
\begin{bmatrix}
    a       & b\\
    c    &   d
\end{bmatrix}
such that a = c = d and b = c - 2, so here is an example
\begin{bmatrix}
    5       & 3\\
    5    &   5
\end{bmatrix}
det(A) = 25 - 15 = 10 , and Tr(A) = 5 + 5 = 10
 A: consider below matrix
$$M=
    \begin{bmatrix}
    a & b \\
    c & d\\
    \end{bmatrix}
$$
the trace will be: 
$$Tr(M)=a+d$$
and the determinant :
$$det(m)=(ad-bc)$$
then according to your problem: 
$$a+d=ad-bc$$
so chose a and d arbitrary and then chose b and c in the way that the above equation will hold. for example:
$$a=10,d=20 $$
bc=170 and you can choose:
$$ b=17 ,c=10$$
A: The condition for the matrix
$$
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
$$
is $a+d=ad-bc$, which becomes, for $a\ne1$,
$$
d=\frac{a+bc}{a-1}
$$
If $a=1$, the condition is $bc=-1$.
Thus you get a family:
$$
\begin{bmatrix}
1 & t \\
-1/t & d
\end{bmatrix}
$$
with arbitrary $t\ne0$ and $d$.
Also
\begin{bmatrix}
a & b \\
c & (a+bc)/(a-1)
\end{bmatrix}
for arbitrary $a\ne 1$, $b$ and $c$.
Note that $d=1$, in the second case, implies $bc=-1$, so it is symmetric with the case $a=1$.
A: If the 2 dimensional matrix $A$ has 2 eigenvalues $x$ and $y$, then $Tr(A)=x+y$ and $det(A)=xy$. So then we have the equation $x+y=xy$ which transforms to:
$$
y = \frac{x}{x-1}
$$
Set some value of $x$, and you'll get a value of $y$. You should then be able to choose any two linearly independent eigenvectors you want. For example, we can choose $x=y=2$, and eigenvectors $\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}$ and then we get your solution of $\begin{pmatrix}2 & 0\\0 & 2\end{pmatrix}$. Another interesting solution is $x=y=0$, which gives the zero matrix. In most cases, however $x$ and $y$ will be unequal, so we might have $x=3, y=3/2$ or $x=-1, y=\frac{1}{2}$.
And the eigenvectors can be as interesting as you like. (As long as they are linearly independent.)
EDIT: As was kindly pointed out to me, to be able to cover non-diagonalizable matrices we need to also allow generalized eigenvectors.
A: Consider an arbitrary $2\times2$ matrix,
$$
A= \begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix}
$$
Its determinant is given by
$$
\det(A)=ad-bc
$$
and its trace is given by
$$
\text{Tr}(A)=a+d
$$
So, we want to know when $\det(A)=\text{Tr}(A)$. That is, when is
$$
ad-bc=a+d
$$
See if you can figure out the conditions on the entries that make the left hand side equal the right hand side. 
