Daily math problems for September 20, 2026 — Analytic Geometry, Algebra, Limits & more

$d=\sqrt{(4-1{)}^2+(6-2{)}^2}=\sqrt{9+16}=5$.

A homomorphism sends the identity to the identity: $\phi (e_G)=e_H$.

$\bullet$ Factor the numerator and cancel:

$$\frac{x^2+x}{x}=\frac{x(x+1)}{x}=x+1(x\mathrm{\ne }0)$$

$\bullet$ Take the limit of the simplified expression:

$$\underset{x\to 0}{\lim }(x+1)=1$$

$(x-3)(x+3)\le 0\iff x\in [-3,3]$. The integers in this interval are $\{-3,-2,-1,0,1,2,3\}$ — seven values.

With $a=2$: $\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{2}{n}})}^n\to e^2$.

By Vieta's relations: $r_1+r_2+r_3=6$ and $r_1{r}_2+r_1{r}_3+r_2{r}_3=11$. So $\sum r_i^2=6^2-2\cdot 11=36-22=14$.

$\sqrt{n+1}-\sqrt{n}={\displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}}}\to 0$.

$b_2=2$, $b_4=8$, $b_6=32$. The sum is $2+8+32=42$. (These three form a GP with first term $2$ and ratio $4$.)

$\bullet$ Evaluate $A(3)$ and its determinant:

$$A(3)=\left(\begin{array}{cc}4& 6\\ 3& -4\end{array}\right)$$

$$\det A(3)=4\cdot (-4)-6\cdot 3=-34$$

$\bullet$ Evaluate $A(2)$ and its determinant:

$$A(2)=\left(\begin{array}{cc}3& 4\\ 3& -5\end{array}\right)$$

$$\det A(2)=3\cdot (-5)-4\cdot 3=-27$$

$\bullet$ $A(3)$ has trace $0$, so by Cayley–Hamilton $A(3{)}^2=34I$. Inverting:

$$A^{-2}(3)=\frac{1}{34}I$$

$\bullet$ Invert $A(2)$:

$$A^{-1}(2)=\frac{1}{27}\left(\begin{array}{cc}5& 4\\ 3& -3\end{array}\right)$$

$\bullet$ Square it:

$$A^{-2}(2)=\frac{1}{729}\left(\begin{array}{cc}37& 8\\ 6& 21\end{array}\right)$$

$\bullet$ One more product:

$$A^{-3}(2)=\frac{1}{19683}\left(\begin{array}{cc}209& 124\\ 93& -39\end{array}\right)$$

$\bullet$ Add the matrices and apply $\det (sI+M)=s^2+s\mathrm{tr}M+\det M$ with $s=\frac{1}{34}$, $\mathrm{tr}M=\frac{170}{19683}$, $\det M=-\frac{1}{19683}$:

$$\det =\frac{1}{1156}+\frac{170}{34\cdot 19683}-\frac{1}{19683}$$

$$=\frac{1}{1156}+\frac{4}{19683}$$

$$=\frac{24307}{1156\cdot 19683}$$

${\displaystyle \frac{1+i}{1-i}}={\displaystyle \frac{(1+i{)}^2}{(1-i)(1+i)}}={\displaystyle \frac{2i}{2}}=i$, whose argument is ${\displaystyle \frac{\pi }{2}}$.