Today's math problems — Euclidean Geometry, Geometry, Matrices & more

$\bullet$ 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:

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

$x-1\to 0^{+}$, so ${\displaystyle \frac{1}{x-1}}\to +\mathrm{\infty }$.

$\bullet$ Evaluate the function values:

$$f(3)=C_3^3+{\log }_{2}4=1+2=3$$

$$f(7)=C_7^3+{\log }_{2}8=35+3=38$$

$\bullet$ So $f(7)\mathrm{/}38=1$ and ${\log }_{\sqrt{3}}f(3)={\log }_{\sqrt{3}}3=2$

$\bullet$ Reduce the trig / limit / log / factorial entries:

$$\tan \frac{\pi }{4}=1$$

$$2^{\lim x^{23}\mathrm{/}x!}=2^0=1$$

$$i^4=1,\lg {10}^{-48}=-48,P_4=4!=24,\ln 1=0$$

$\bullet$ Assemble the matrix:

$$A=\left(\begin{array}{ccc}3& 1& 1\\ 1& 2& 1\\ -48& 24& 0\end{array}\right)$$

$\bullet$ Expand along row 1:

$$\det A=3(2\cdot 0-1\cdot 24)-1(1\cdot 0-1\cdot (-48))+1(1\cdot 24-2\cdot (-48))$$

$$\det A=3(-24)-1(48)+1(120)$$

$$\det A=-72-48+120=0$$