Daily Math · 2026-05-13

Question 1

For $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & -1 \ 5 & 2 \end{pmatrix}$, $A + B$ equals:

Accepted Answer

Add componentwise: $\begin{pmatrix} 1 & 1 \ 8 & 6 \end{pmatrix}$.

Answer

$\begin{pmatrix} 1 & 1 \ 8 & 6 \end{pmatrix}$

Answer

$\begin{pmatrix} 1 & 3 \ 8 & 2 \end{pmatrix}$

Answer

$\begin{pmatrix} 1 & -1 \ -2 & 2 \end{pmatrix}$

Answer

$\begin{pmatrix} 0 & -2 \ 15 & 8 \end{pmatrix}$

Question 2

Determine $n \in \mathbb{N}$, $n \ge 2$, such that $\binom{n}{2} = 15$.

Accepted Answer

$\dfrac{n(n-1)}{2} = 15 \Rightarrow n^2 - n - 30 = 0 \Rightarrow n \in \{6, -5\}$. Only $n = 6$ is admissible.

Answer

$4$

Answer

$5$

Answer

$6$

Answer

$15$

Question 3

The solution set of the inequality $\log_{1/3}(x^2 - 4) > 0$ is:

Accepted Answer

**Domain constraint:** $x^2 - 4 > 0 \Rightarrow x \in (-\infty, -2) \cup (2, +\infty)$. **Inequality:** $\log_{1/3}(x^2 - 4) > 0 = \log_{1/3} 1$. Since the base $\dfrac{1}{3} < 1$, $\log_{1/3}$ is decreasing, so the inequality flips: $x^2 - 4 < 1$, i.e. $x^2 < 5$, so $x \in (-\sqrt{5}, \sqrt{5})$. **Intersection with the domain:** $\left[(-\infty, -2) \cup (2, +\infty)\right] \cap (-\sqrt{5}, \sqrt{5}) = (-\sqrt{5}, -2) \cup (2, \sqrt{5})$.

Answer

$x \in (-\infty, 0)$

Answer

$x \in (0, 2)$

Answer

$x \in (-\sqrt{5}, -2) \cup (2, \sqrt{5})$

Answer

$x \in (\sqrt{5}, \infty)$

Answer

$x \in (2, \infty)$

Answer

$x \in (0, \sqrt{5})$

Problems & worked solutions

Problem 1 — Matrices

Problem 2 — Permutations & Combinations

Problem 3 — Logs