Daily math problems for August 28, 2026 — Matrices, Calculus, Algebra & more

$\bullet$ Compute $A^2$ to spot the pattern:

$$A^2=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$$

$\bullet$ By induction:

$$A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$$

$\bullet$ Therefore:

$$A^{2026}=\left(\begin{array}{cc}1& 2026\\ 0& 1\end{array}\right)$$

$\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{2}}\right)}^n\to 0$ since $\mathrm{\mid }q\mathrm{\mid }={\displaystyle \frac{1}{2}}<1$.

$\bullet$ Geometric series with first term $a=2$, ratio $r=2$, $n=10$ terms:

$$S=a\cdot {\displaystyle \frac{r^n-1}{r-1}}$$

$\bullet$ Substitute:

$$S=2\cdot {\displaystyle \frac{2^{10}-1}{1}}$$

$\bullet$ Evaluate:

$$S=2\cdot 1023=2046$$

The polynomial $z^n-1$ has zero coefficient for $z^{n-1}$, so by Viète's formula the sum of its roots is $0$. Equivalently, the geometric sum ${\sum }_{k=0}^{n-1}{e}^{2\pi ik\mathrm{/}n}={\displaystyle \frac{1-1}{1-e^{2\pi i\mathrm{/}n}}}=0$ (for $n\ge 2$).

$A\mathrm{△}\mathrm{\varnothing }=A$, so $\mathrm{\varnothing }$ is the identity.

${\lim }_{x\to \mathrm{\infty }}{e}^{-x}=0$, so $y=0$ is a horizontal asymptote at $+\mathrm{\infty }$.

Negation of $\mathrm{\forall }$ is $\mathrm{\exists }$; negation of $>$ is $\le$. So the negation is $\mathrm{\exists }x\in \mathbb{R}:x^2+1\le 0$. (This statement is itself false, but the question asked for the negation, not its truth value.)

$4^x=2^{2x}=(2^x{)}^2=5^2=25$.

$P(2)=8-2m+4=12-2m=0\Rightarrow m=6$.

A linear system is consistent (has at least one solution) iff $\mathrm{rank}(A)=\mathrm{rank}([A\mathrm{\mid }b])$ — the rank of the coefficient matrix equals the rank of the augmented matrix.