Daily math problems for July 3, 2026 — Matrices, Calculus

$\bullet$ Expand the determinant by row reduction or cofactors:

$$\mid \begin{array}{ccc}1& 2& 3\\ 2& x+1& 2x+1\\ 3& 2x+1& 3x+1\end{array}\mid =-x^2+3x-1$$

$\bullet$ Set it equal to $-1$:

$$-x^2+3x-1=-1⟺x(x-3)=0$$

$\bullet$ The two roots are $0$ and $3$, so their sum is:

$$0+3=3$$

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t\in [7,17]$ - $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7,17]$ (a hiker doesn't teleport).

Now consider $f(t)=u(t)-d(t)$. By the problem statement:

- $u(7)=0$ and $d(7)=L$, so $f(7)=-L<0$. - $u(17)=L$ and $d(17)=0$, so $f(17)=+L>0$.

Since $f$ is continuous and changes sign on $[7,17]$, the Intermediate Value Theorem guarantees some $t^{\ast }\in (7,17)$ with $f(t^{\ast })=0$, i.e. $u(t^{\ast })=d(t^{\ast })$. At that clock time, the monk stands at the same point on the trail on both days. $\mathrm{■}$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

By induction, $A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, so $(A^n{)}_{12}=n$.