Daily math problems for September 10, 2026 — Matrices, Financial Mathematics, Elementary Functions & more

$\bullet$ Compute $\det A$ by cofactor expansion along row $1$:

$$\det A=0\cdot M_{11}-(-1)\cdot M_{12}+3\cdot M_{13}=2+3=5$$

$\bullet$ For an $n\times n$ matrix, $\det (cM)=c^n\det M$ and $\det (M^{-1})=1\mathrm{/}\det M$:

$$\det (5A^{-1})=5^3\cdot \frac{1}{\det A}=\frac{125}{5}=25$$

$I=1000\cdot 0.04\cdot 5=200$.

$f(1)=1$, $f(2)={\displaystyle \frac{1}{2}}$. As $x$ grows, ${\displaystyle \frac{1}{x}}$ shrinks — $f$ is strictly decreasing on $(0,+\mathrm{\infty })$.

$\bullet$ Evaluate each special entry:

$$i^2=-1,\lg 100=2,\ln 1=0$$

$$\sin \frac{\pi }{2}=1,i^{24}=1,C_4^3=4$$

$\bullet$ Assemble the numerical matrix:

$$A=\left(\begin{array}{cccc}1& 0& 2& 4\\ -1& 1& 3& 2\\ 0& 1& 3& 0\\ 1& 1& 4& 7\end{array}\right)$$

$\bullet$ Row-reduce. $R_2+=R_1$, $R_4-=R_1$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 1& 3& 0\\ 0& 1& 2& 3\end{array}\right)$$

$\bullet$ $R_3-=R_2$, $R_4-=R_2$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 0& -2& -6\\ 0& 0& -3& -3\end{array}\right)$$

$\bullet$ $R_4-=\frac{3}{2}{R}_3$ gives upper-triangular with pivots $1,1,-2,6$:

$$\det A=1\cdot 1\cdot (-2)\cdot 6=-12$$

Since $5^2+{12}^2={13}^2$, $\mathrm{△}ABC$ is right-angled at $B$. Area $={\displaystyle \frac{1}{2}}\cdot 5\cdot 12=30$.

$f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$. $f^{\mathrm{\prime }\mathrm{\prime }}<0\iff x<0$. So $f$ is concave on $(-\mathrm{\infty },0)$.

Associative ✓, identity $1$ ✓, every element has an inverse ✓. Hence a (commutative) group.

$\int \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+1}})dx=\ln \mathrm{\mid }x\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid }+C=\ln \text{\hspace{0.17em}⁣}\mid {\displaystyle \frac{x}{x+1}}\mid +C$.

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{x}}=2\cdot \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{2x}}=2\cdot 1=2}$. So $a=2$.

$g(0)=1>0$, $g(\pi \mathrm{/}2)=0-\pi \mathrm{/}2<0$. Since $g$ is continuous, by the IVT there is $c\in (0,\pi \mathrm{/}2)$ with $g(c)=0$, i.e. $\cos c=c$.