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I am confused about this homework question.
It says "Show that :
$\det(A) = \frac 12 \begin{vmatrix}\operatorname{tr}(A)&1\\\operatorname{tr}(A^2)& \operatorname{tr}(A)\end{vmatrix}$
for every $2\times 2$ matrix."
I am not sure how to "show" this. I'm guessing it means a proof of sorts.
But I am also confused as to why the $\det(A) = 1/2$ times a matrix. I'm pretty sure determinants are just a number, not a matrix.
Any tips would be great. Thanks in advance.
The "|" to the left and the right of your matrix probably indicate that you should take the determinant of this matrix, too. It is not completly uncommon to write $|A|$ instead of $det(A)$.
To prove the identity just take an abstract $2 \times 2$-Matrix and compute the left-hand and the right-hand side of the equation.
Here is a (requested) generalization for larger dimensions: If A is an n×n matrix, then one has the following expression for the determinant of A in terms of the determinant of a matrix whose entries are traces of powers of A:
$$\newcommand{\tr}{\operatorname{tr}} \newcommand{\det}{\operatorname{det}} \det(A) = \frac{1}{n!}\left|\begin{array}{cccccccc} \tr(A) & 1 & . & . & . & \dots & . & . \\ \tr(A^2) & \tr(A) & 2 & . & . & \dots & . & . \\ \tr(A^3) & \tr(A^2) & \tr(A) & 3 & . & \dots & . & . \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \tr(A^{n-5}) & \dots & \tr(A) & n-1 \\ \tr(A^n) & \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \dots & \tr(A^2) & \tr(A) \end{array}\right|$$
In particular, we have for $3\times3$ that: $$\det(A) = \frac{1}{6}\left|\begin{array}{cccc} \tr(A) & 1 & 0 \\ \tr(A^2) & \tr(A) & 2 \\ \tr(A^3) & \tr(A^2) & \tr(A) \end{array}\right|$$
The matrix on the right is defined in general by: $$B_{ij} = \begin{cases} i & \text{ if } j = i + 1 \\ \tr(A^{i-j+1}) & \text{ if } j \leq i \\ 0 & \text{ if } j > i+1 \end{cases}$$
I suggest using this recursively to create even more complicated formulas (all in lower Hessenberg form!) to be entered into the big book of bad algorithms.
Assuming you are familiar with Cayley–Hamilton theorem, this may be an alternative way:
Any $2\times2$ matrix satisfies $A^2-\text{tr}(A)A+\det(A)I_2=0 $. Taking trace of both sides, $\begin{align*}&\text{tr}(A^2)-\text{tr}(A)\text{tr}(A)+2\det(A)=0 \Rightarrow\\ &\det(A)=\frac12[\text{tr}(A)\text{tr}(A)-\text{tr}(A^2)]=\frac12\begin{vmatrix}\text{tr}(A)&1\\\text{tr}(A^2)&\text{tr}(A)\end{vmatrix}\end{align*}$
You could use an eigenvalue formulation, since if $\lambda_1,\lambda_2$ are the eigenvalues of $A$ then $\det(A)=\lambda_1 \lambda_2$, $tr A=\lambda_1+\lambda_2$, $tr(A^2)=\lambda_1^2+\lambda_2^2$. Plug these in your equation and obvious equality holds.
The RHS is determinant, i.e., $1/2[(tr A)^2-tr(A^2)]=1/2[(a_{11}+a_{22})^2-(a_{11}^2+2a_{12}a_{21}+a_{22}^2)]=\cdots$
