# 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

• So far, you have one example. You need two. But you have only explored matrices of the form $aI$. There are other matrices too.... Commented Aug 10, 2018 at 5:11
• @LordSharktheUnknown, I am still thinking and trying other matrices other than aI that would work. I haven't found any yet. Could you please let me know what kind of matrix form I can look into? I appreciate your hint :) Commented Aug 10, 2018 at 5:13
• Do you know what eigenvalues are, and the relationship between eigenvalues, the trace and the determinant? If not, then I would suggest that you write down the entries of a $2 \times 2$ matrix as variables. You know the trace is sum of the diagonal entries, and the determinant formula. Equating both of these gives you one equation in four variables : you should be able to find some non-trivial solutions. Commented Aug 10, 2018 at 5:14
• That is to say : you do not know that the trace is the sum of all the eigenvalues, and the determinant is the product of all the eigenvalues? If so, then the approach via variables should work. Commented Aug 10, 2018 at 5:16
• Sorry I meant I know how trace and determinant is calculated, excuse my English :) . I am trying your suggestion right now Commented Aug 10, 2018 at 5:18

consider below matrix $$M= \begin{bmatrix} a & b \\ c & d\\ \end{bmatrix}$$ the trace will be: $$Tr(M)=a+d$$

and the determinant : $$det(m)=(ad-bc)$$

$$a+d=ad-bc$$

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: $$b=17 ,c=10$$

• I appreciate your explanation and example. Based on some trial and error, I realized the Tr(A) = det(A) will work when a = d = c and b = c - 2 while eigenvalue lambda is 0. Am I right? Commented Aug 10, 2018 at 20:02
• I just added my new solution to my question, could you please check that for me? Would my answer be acceptable? Commented Aug 11, 2018 at 1:12

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 $a=1$.

• I appreciate your explanation, it is an interesting way to drive it. I think there might be an easier way to do this, I have added my new solution to my question (look for the Edited section). Could you please have a look at it, and let me know if it is an acceptable answer? Commented Aug 11, 2018 at 1:12

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.

• Eigenvectors can be as interesting as you like but have to be orthogonal. Commented Aug 11, 2018 at 7:36
• I tried it with eigenvectors <1, 2>, <1, -2> and eigenvalues 1/2, -1, and it worked fine. Did you mean linearly independent? I'll edit to specify that the eigenvectors are l.i. Commented Aug 14, 2018 at 0:28
• Yes linear independent. But also take into account that this solution does not cover all cases. Here you cover all diagonalizable matrices. What about the other ones? Commented Aug 14, 2018 at 6:32
• You're right. I guess I should say $generalized$ eigenvectors. Commented Aug 14, 2018 at 23:00

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.

• I appreciate your explanation and example. Based on some trial and error, I realized the Tr(A) = det(A) will work when a = d = c and b = c - 2 while eigenvalue lambda is 0. Am I right? Commented Aug 10, 2018 at 19:59
• I added my new solution to my question, could you please check it for me? Commented Aug 11, 2018 at 1:13