heureka

We're going to consider the matrix

\(\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\)

a)
Let \({P} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}\).
Find the 2x2 matrix D such that

\({P} {D} {P}^{-1} = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\)

\(\text{Let $M = \begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix} $ }\)

Formula:

\(\begin{array}{|rcll|} \hline P^{-1} &=& \dfrac{1}{\begin{vmatrix} 3 & -5 \\ -1 & 2 \end{vmatrix}}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{6-5}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ &=& \dfrac{1}{1}\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \\\\ \mathbf{P^{-1}} & \mathbf{=} & \mathbf{\begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline PDP^{-1} &=& M \quad | \quad \cdot P \\ PDP^{-1}P &=& MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ PDI &=& MP \quad | \quad \cdot P^{-1} \\ P^{-1}PDI &=& P^{-1}MP \quad | \quad P^{-1}P = I(\text{Identity matrix}) \\ IDI &=& P^{-1}MP \\ \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{D} & \mathbf{=} & \mathbf{P^{-1}MP} \\ &=& \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}\begin{pmatrix} -4 & -15 \\ 2 & 7 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ &=& \begin{pmatrix} 2 & 5 \\ 2 & 6 \end{pmatrix}\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \\ \mathbf{D} & \mathbf{=} & \mathbf{\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}} \quad | \quad (\text{Diagonal matrix}) \\ \hline \end{array}\)

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. 

x^5y^2-x^3y+5xy^3=0

\(\begin{array}{|l|rcll|} \hline & \mathbf{x^5y^2-x^3y+5xy^3} & \mathbf{=}& \mathbf{0} \\\\ \hline y'=\ ? & (5x^{5-1}\cdot y^2+ x^5\cdot 2y^{2-1}\cdot y') \\ & - (3x^{3-1}\cdot y+ x^3\cdot y') \\ & + 5(1\cdot x^{1-1}\cdot y^3 +x\cdot 3y^{3-1}\cdot y' ) &=& 0 \\ \hline & (5x^4y^2+ x^5\cdot 2yy') - (3x^2y+ x^3y') + 5(y^3 +x\cdot 3y^2y' ) &=& 0 \\ & 5x^4y^2+ 2x^5yy'- 3x^2y - x^3y' + 5y^3 +15xy^2y' &=& 0 \\ & 2x^5yy'- x^3y'+15xy^2y'&=& 3x^2y-5y^3-5x^4y^2 \\ & y'x(2x^4y-x^2+15y^2)&=& y(3x^2-5y^2-5x^4y) \\ & \mathbf{y'} &\mathbf{=}& \mathbf{\dfrac{y(3x^2-5y^2-5x^4y)}{x(2x^4y-x^2+15y^2)}} \\ \hline \end{array}\)

14+14-14*14/14+15+15-15*15/15+3+3-3*3/3

\(\begin{array}{|rcll|} \hline && 14+14-14*\underbrace{\dfrac{14}{14}}_{=1}+15+15-15*\underbrace{\dfrac{15}{15}}_{=1}+3+3-3*\underbrace{\dfrac{3}{3}}_{=1} \\ &=& 14+14-14*1+15+15-15*1+3+3-3*1 \\ &=& 14+\underbrace{14-14}_{=0} +15+\underbrace{15-15}_{=0} +3+\underbrace{3-3}_{=0} \\ &=& 14+0 +15+0 +3+0 \\ &=& \mathbf{32} \\ \hline \end{array}\)

In right triangle \(XYZ\) with \(\angle YXZ = 90^\circ\), we have \(XY = 24\) and \(YZ = 25\). Find \(\tan Y\).

\(\begin{array}{|rcll|} \hline 24^2+XZ^2 &=& 25^2 \\ XZ^2 &=& 25^2-24^2 \\ &=& (24+1)^2 - 24^2 \\ &=& 24^2 + 48 + 1 - 24^2 \\ &=& 49 \\ &=& 7^2 \\ \mathbf{XZ} &\mathbf{=} & \mathbf{7} \\\\ \tan(Y) &=& \dfrac{XZ}{24} \\ \mathbf{\tan(Y)} &\mathbf{=} & \mathbf{\dfrac{7}{24}} \\ \hline \end{array}\)

Find y'' by implicit differentiation.

5x^3 + 4y^3 = 6

\(\begin{array}{|l|rcll|} \hline & \mathbf{5x^3 + 4y^3} & \mathbf{=}& \mathbf{6} \\\\ \hline y'=\ ? & 5\cdot 3x^{3-1} + 4\cdot 3y^{3-1}\cdot y' &=& 0 \\ & 15x^2+12y^2y' &=& 0 \\ & \mathbf{y'} &\mathbf{=}& \mathbf{-\dfrac{5x^2}{4y^2}} \\ \hline y''=\ ? & 15x^2+12y^2y' &=& 0 \\ & 15\cdot2 x^{2-1}+12\cdot\left( (y^2)'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x^{2-1}+12\cdot\left( 2\cdot y^{2-1}y'\cdot y' +y^2\cdot y'' \right) &=& 0 \\ & 30 x+12\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \quad |\quad : 6 \\ & 5x+ 2\cdot\left( 2\cdot y(y')^2 +y^2\cdot y'' \right) &=& 0 \\ & 5x+ 4y(y')^2 + 2y^2\cdot y'' &=& 0 \quad |\quad \mathbf{y'=-\dfrac{5x^2}{4y^2}}\\ & 5x+ 4y\left(-\dfrac{5x^2}{4y^2}\right)^2 + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{4y\cdot 25x^4}{16y^4} + 2y^2\cdot y'' &=& 0 \\ & 5x+ \dfrac{25x^4}{4y^3} + 2y^2\cdot y'' &=& 0 \quad |\quad \cdot 4y^3 \\ & 20xy^3+ 25x^4 + 8y^5\cdot y'' &=& 0 \\ & 5x(\underbrace{4y^3+ 5x^3}_{=6}) + 8y^5\cdot y'' &=& 0 \\ & 5x\cdot 6 + 8y^5\cdot y'' &=& 0 \\ & 30x + 8y^5\cdot y'' &=& 0 \quad |\quad :2 \\ & 15x + 4y^5\cdot y'' &=& 0\\ & 4y^5\cdot y'' &=& -15x \quad |\quad :4y^5 \\ & \mathbf{y''} &\mathbf{=}& \mathbf{-\dfrac{15x}{4y^5}} \\ \hline \end{array}\)

The equation of the circle that passes through (-1,6) and which has a center at (2,3)
can be written as x^2 + y^2 + Ax + By + C = 0.
Find A*B*C

\(\text{Let $x_c=2,\ y_c =3,\ x_p=-1,\ y_p=6$} \)

\(\begin{array}{|rcll|} \hline (x-x_c)^2+(y-y_c)^2 &=& r^2 \quad | \quad r^2 = (x_p-x_c)^2+(y_p-y_c)^2 \\ (x-x_c)^2+(y-y_c)^2 &=& (x_p-x_c)^2+(y_p-y_c)^2 \\ x^2-2xx_c+ \not{x_c^2}+y^2-2yy_c+\not{y_c^2} &=& x_p^2-2x_px_c+\not{x_c^2}+y_p^2-2y_py_c+\not{y_c^2} \\ x^2-2xx_c +y^2-2yy_c &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+(-2x_c)x+(-2y_c)y &=& x_p^2-2x_px_c +y_p^2-2y_py_c \\ x^2+y^2+ (-2x_c) x+ (-2y_c)y -(x_p^2-2x_px_c +y_p^2-2y_py_c ) &=& 0 \\ x^2+y^2+\underbrace{(-2x_c)}_{=A}x+\underbrace{(-2y_c)}_{=B}y +\underbrace{(2x_px_c+2y_py_c -x_p^2-y_p^2)}_{=C} &=& 0 \\\\ A&=& -2x_c\\ &=& -2\cdot 2 \\ \mathbf{A} &=& \mathbf{-4} \\\\ B&=& -2y_c\\ &=& -2\cdot 3 \\ \mathbf{B} &=& \mathbf{-6} \\\\ C &=& 2x_px_c+2y_py_c -x_p^2-y_p^2 \\ &=& 2(-1)2+2\cdot 6 \cdot 3-(-1)^2-(6)^2 \\ &=& -4+36-1-36 \\ \mathbf{C} &=& \mathbf{-5} \\\\ A\cdot B\ \cdot C &=& (-4)(-6)(-5) \\ &=& -4\cdot 5 \cdot 6 \\ \mathbf{A\cdot B \cdot C } &=& \mathbf{-120} \\ \hline \end{array} \)

Thank you, JM

I have tried this sum 3 times and got 3 different answers, please help me with this one:

\(\mathbf{\large{ \dfrac{ 54^n-18^{n-1}\cdot 3^{n+1} } { {(3^{n-1})}^2\cdot 6^n } }}\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{ 54^n-18^{n-1}\cdot 3^{n+1} } { {(3^{n-1})}^2\cdot 6^n } } \\\\ &=& \dfrac{ 54^n-18^n\cdot 18^{-1}\cdot 3^n\cdot 3^1 } { { 3^{(n-1)\cdot 2} } \cdot 6^n } \\\\ &=& \dfrac{ 54^n-18^n\cdot 3^n\cdot \frac{3}{18}} { {3^{2n-2}}\cdot 6^n } \\\\ &=& \dfrac{ 54^n-(18\cdot 3)^n\cdot \frac{1}{6} } { {3^{2n}\cdot 3^{-2} }\cdot 6^n } \\\\ &=& \dfrac{ 54^n-54^n\cdot \frac{1}{6} } { \left(3^2\right)^n\cdot \frac{1}{9} \cdot 6^n } \\\\ &=& \dfrac{ 54^n \cdot \left(1- \frac{1}{6}\right) } { 9^n\cdot 6^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ 54^n \cdot \frac{5}{6} } { \left(6\cdot 9\right)^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ 54^n \cdot \frac{5}{6} } { 54^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ \frac{5}{6} } { \frac{1}{9} } \\\\ &=& \dfrac{5}{6} \cdot \dfrac{9}{1} \\\\ &=& \dfrac{5\cdot 3}{2} \\\\ &\mathbf{=}& \mathbf{\dfrac{15}{2}} \\ \hline \end{array}\)

T1=2, T2=6 and T3=12. Determine the nth term formula.

T1=2, T2=6 and T3=12. Determine the nth term formula.

\(\begin{array}{|rcll|} \hline t_1 = 1 \cdot 2 &=& 2 \\ t_2 = 2 \cdot 3 &=& 6 \\ t_3 = 3 \cdot 4 &=& 12 \\ \ldots \\ t_n &=& n(n+1) \\ \hline 420 &=& n(n+1) \\ 420 &=& n^2+n \\ n^2+n -420 &=& 0 \\ n&=& \dfrac{-1\pm\sqrt{1-4\cdot(-420)}}{2} \\ n&=& \dfrac{-1\pm\sqrt{1681}}{2} \\ n&=& \dfrac{-1\pm 41}{2} \\ n&=& \dfrac{-1 {\color{red}+} 41}{2} \quad | \quad n > 0\ ! \\ \mathbf{n} &\mathbf{=}& \mathbf{ 20 }\\ \hline \end{array}\)

\(\mathbf{t_{20} = 20\cdot 21 = 420}\)