Tomi's Challenges - #5

Algebra Level 4

Let $g(A) : A\rightarrow A^T$ , $g^{n}(A)=(g\circ g\circ g\circ\cdots\circ g)(A)$ , $n$ times, and let $A$ be the following matrix:

$A=\begin{bmatrix} x^{e}\cdot sin(x) & y^{4}\cdot \text{sgn}(\sin(2021y^3)) & -y\\ 8y & yx^{90}-x^2+91 & 1\\ x^2-(y+1)x & \text{sgn}(y) & 82\sqrt{53}y \end{bmatrix}$

Find the sum of the pair $(x, y)$ , namely $x+y$ , such that $\lim \limits_{n\to\infty} g^n(A)$ is defined, if $x, y$ are unique. If there are multiple solutions for $x$ and $y$ , input the sum of sums of all such pairs.

Notation: $\text{sgn}(x) := \begin{cases} \begin{array} {l l } -1 & \text{ if }x<0 \\ 0 & \text{ if }x=0 \\ 1 & \text{ if }x>0 \\ \end{array} \end{cases}$ denotes the sign function.

The answer is 7.

1 solution

Tomislav Franov
Apr 17, 2021

If we transpose a matrix, we flip it over its main diagonal. If we transpose it one more time, we return it to its initial state. In general, transposing a matrix will always yield a different matrix unless that matrix is symmetrical. So if we transpose a matrix a very large number of times, we can't be certain whether it will land at its initial state or at its transposed state, again, unless it's symmetrical - because then the limit is just the identity operator - i.e. it does nothing to the matrix. So, the takeaway is that we need the matrix to be symmetrical for this limit to be well defined.

This means that the following three equations should be satisfied:

$\text{sgn}(y)=1$
$x^2-(y+1)x=-y$
$8y=y^4\cdot \text{sgn}(\sin(2021y^3))$

From $sgn(y)=1$ we have $y>0$ . The logical next step is to solve for y in the third equation. Taking the absolute value on both sides, we get:

$|8y|=|y^4 \cdot \text{sgn}(\sin(2021y^3))|$
$|8y|=|y^4|$ - since $\sin(2021y^3)$ is an odd function.
$8y=y^4, -8y=y^4$

After solving these equations we get $y\in{{-2, 0, 2}}$ , and since $y>0$ , we get that $y=2$ . Now we solve for $x$ in the 2nd equation:

$x^2-3x=-2$

After solving this we get $x\in{{2, 1}}$ . So the solutions are $(x,y)={(1,2), (2,2)}$ . So the solution is $1+2+2+2=\boxed{7}$ .

Amazing problem! Shame for me that I don't know matrices to be able to solve it. Hope you keep uploading more!

Veselin Dimov - 1 month, 3 weeks ago

Thanks :D I will keep uploading, I just hope I'll be able to come up with these problems frequently enough..

Tomislav Franov - 1 month, 3 weeks ago

Tomi's Challenges - #5

The answer is 7.

1 solution

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