Daily math problems for September 16, 2026 — Algebra, Matrices, Functions & more

An isomorphism is a bijective homomorphism. (Equivalently, a homomorphism whose inverse is also a homomorphism.)

$\bullet$ Compute the determinant:

$$\mid \begin{array}{cc}x& 3x+1\\ -2& 7\end{array}\mid =7x-(-2)(3x+1)=13x+2$$

$\bullet$ Expand the arrangement (for $x\ge 2$):

$$A_x^2=\frac{x!}{(x-2)!}=x(x-1)$$

$\bullet$ The equation becomes:

$$13x+2+x(x-1)=0$$

$$x^2+12x+2=0$$

$\bullet$ By Vieta's: $x_1+x_2=-12$ and $x_1{x}_2=2$.

$\bullet$ Sum of squares:

$$x_1^2+x_2^2=(-12{)}^2-2\cdot 2=140$$

The graph of $f^{-1}$ is the reflection of the graph of $f$ across $y=x$. Their intersection lies on $y=x$, so we solve $f(x)=x$: $2x-3=x\Rightarrow x=3$.

The integral of a non-negative continuous function over $[a,b]$ is non-negative (representing the area under the curve).

$\bullet$ Apply $\ln (1+u)\sim u$ when $u\to 0$:

$$\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{2^n})\sim \frac{1}{2^n}$$

$\bullet$ The product reduces to a geometric ratio:

$$3^n\cdot \frac{1}{2^n}={\left(\frac{3}{2}\right)}^n$$

$\bullet$ Since $3\mathrm{/}2>1$, this diverges:

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{3}{2}\right)}^n=+\mathrm{\infty }$$

$5$ is prime, so ${\mathbb{Z}}_5$ is a field. (Every non-zero element has a multiplicative inverse.)

The second equation is $2\times$ the first, so it provides no new information. The system has infinitely many solutions.

With $Y=X^2$: $Y^2-5Y+4=0\Rightarrow Y\in \{1,4\}$. Then $X^2=1\Rightarrow X=\pm 1$ and $X^2=4\Rightarrow X=\pm 2$. Four real solutions.

$R=\sqrt{1+3}=2$, so $f(x)=2\sin (x+\pi \mathrm{/}3)$. Max value: $2$.

$P(-1)=0$, so $(X+1)$ is a factor. Polynomial division: $P(X)=(X+1)(X^2-4)=(X+1)(X-2)(X+2)$.