2019 Matriculation Examination
Sample Paper
Mathematics                              Time allowed : 3 hours
Section (A)

1. (a) Let the function $\displaystyle f(x)=\frac{{1-2x}}{{1+x}},x\ne 1$. If $\displaystyle {{g}^{{-1}}}(x)={{f}^{{-1}}}(x+1)$, evaluate $\displaystyle g(2)$.
(3 marks)
(b) The expression $\displaystyle (x + 4)^3 + ax + b$ has a factor $\displaystyle x + 1$ but leaves a remainder of $\displaystyle 8$ when divided by $\displaystyle x + 5$. Find the values of $\displaystyle a$ and $\displaystyle b$.
(3 marks)
2.  (a) If the first four terms in the expansion of $\displaystyle (x^2−2)^5$ in descending powers of $\displaystyle x$ are $\displaystyle x^{10}−10x^8+40x^6+Ax^4+...$, find the value of $\displaystyle A$.
(3 marks)
(b) In a geometric progression, $\displaystyle {{u}_{1}}=\frac{1}{{81}}$ and $\displaystyle {{u}_{4}}=\frac{1}{{3}}$. Find the common ratio.
(3 marks)
3  (a) Find the two matrices of the form $\displaystyle P=\left( {\begin{array}{*{20}{c}} 4 \\ {x-2} \end{array}\ \ \ \begin{array}{*{20}{c}} {{{x}^{2}}-2x} \\ {-1} \end{array}} \right)$ such that $\displaystyle P=P'$.
(3 marks)
(b) A fair coin is tossed 5 times. What is the probability of getting at least one head?
(3 marks)
4.  (a) In the figure, $\displaystyle ∠ABC=30°$, $\displaystyle AB=BC$ and $\displaystyle AD$ is a tangent. Find $\displaystyle ∠BDA$.
(3 marks)
(b)  $\displaystyle A,B$, and $\displaystyle C$ are with position vectors $\displaystyle \hat{\text{i}}+3\hat{\text{j}}$, $\displaystyle 2\hat{\text{i}}+5\hat{\text{j}}$ and $\displaystyle k\hat{\text{i}}-4\hat{\text{j}}$  respectively. Find the value of $\displaystyle k$ if $\displaystyle A,B$, and $\displaystyle C$ are collinear.
(3 marks)
5.  (a) Prove that $\displaystyle {{(1-\tan x)}^{2}}+{{(1-\cot x)}^{2}}={{(\sec x-\operatorname{cosec}x)}^{2}}$.
(3 marks)
(b) Evaluate (i) $\displaystyle \underset{{x\to 1}}{\mathop{{\lim }}}\,\frac{{\sqrt[3]{x}-1}}{{x-1}}$   (ii) $\displaystyle \underset{{x\to \pi }}{\mathop{{\lim }}}\,\frac{{\cos \frac{x}{2}}}{{\pi -x}}$.
(3 marks)
Section (B)

6.    (a) A function $\displaystyle f$ is defined by $\displaystyle f:x\mapsto \frac{x}{p}+q,p\ne 0$. If $\displaystyle f(8)=1$ and$\displaystyle {{f}^{{-1}}}(-2)=2$, show that $\displaystyle \frac{p}{2}+{{q}^{2}}=10$.
(5 marks)
(b)  Given that the $\displaystyle (p + 1)^{\text{th}}$ term of an A.P. is twice the $\displaystyle (q + 1)^{\text{th}}$ term. Prove that$\displaystyle (3p + 1)^{\text{th}}$ term is twice the $\displaystyle (p +q+ 1)^{\text{th}}$ term.
(5 marks)
7.    (a) Find the term in $\displaystyle x^2$ and $\displaystyle x^3$ in the expansion of  $\displaystyle(2x+1)^5$. Hence find the term in $\displaystyle x^3$ in the expansion of  $\displaystyle (x+3)(2x+1)^5$.
(5 marks)
(b) Let $\displaystyle {{\text{J}}^{+}}$ be the set of all positive integers. Is the operation $\displaystyle \odot$ defined by $\displaystyle x\odot y=x^2+3y$ a binary operation on  $\displaystyle {{\text{J}}^{+}}$? If it is a binary operation, solve the equation $\displaystyle \left( {k\odot 5} \right)-\left( {3\odot k} \right)=3k+1$.
(5 marks)
8.    (a) If $\displaystyle k+4, k$ and $\displaystyle 2k-15$ where $\displaystyle k>0$ are the first three terms of a geometric progression, find the value of $\displaystyle k$. Hence find the first term and the common ratio and determine whether the sum to infinity exists or not. Find the sum to infinity of the progression if exists.
(5 marks)
(b) Find the solution set in $\displaystyle \text{R}$ of the in equations $\displaystyle (x+2)^2>2x+7$ and illustrate it on the number line.
(5 marks)
9.   (a) Given that when $\displaystyle f(x)=6x^3+3x^2+ax+b$, where $\displaystyle a$ and $\displaystyle b$ are constants, is divided by $\displaystyle (x+1)$ the remainder is $\displaystyle 45$, show that $\displaystyle b – a=48$. Given also that $\displaystyle (2x+1)$ is a factor of $\displaystyle f(x)$, find the value of $\displaystyle a$ and of $\displaystyle b$. Hence factorise $\displaystyle f(x)$ completely.
(5 marks)
(b) If $\displaystyle A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)$ show that $\displaystyle A+{{A}^{{-1}}}-2I=O$ where $\displaystyle I$ is a unit matrix of order 2.
(5 marks)
10.  (a) The matrices $\displaystyle P,Q$ and $\displaystyle R$ such that$\displaystyle P=\left( {\begin{array}{*{20}{c}} 2 & 1 \\ 3 & 2 \end{array}} \right)$, $\displaystyle Q=\left( {\begin{array}{*{20}{c}} {-1} & 0 \\ 2 & 1 \end{array}} \right)$ and $\displaystyle R=PQ$. Verify that $\displaystyle {{Q}^{{-1}}}{{P}^{{-1}}}={{R}^{{-1}}}$.
(5 marks)
(b) The probability that a student will receive an $\displaystyle A, B, C$ or $\displaystyle D$ grade are 0.3, 0.38, 0.22 and 0.1 respectively. What is the probability that student will receive
(i) at least $\displaystyle B$ grade?
(ii) at most $\displaystyle C$ grade?
(iii) not an $\displaystyle A$ grade?
(iv) $\displaystyle B$ or $\displaystyle C$ grade?
(5 marks)
Section (C)

11.   (a) Two unequal circles are tangent externally at $\displaystyle O$. $\displaystyle AB$ the chord of the first circle is tangent to the second circle at $\displaystyle C$, and $\displaystyle AO$ meets this circle at $\displaystyle E$. Prove that $\displaystyle ∠BOC=∠COE$.
(5 marks)
(b) In $\displaystyle ΔABC$, $\displaystyle D$ is a point of $\displaystyle AC$ such that $\displaystyle AD=2CD$. $\displaystyle E$ is on $\displaystyle BC$ such that $\displaystyle DE \parallel AB$. Compare the areas of $\displaystyle ΔCDE$ and $\displaystyle ΔABC$. If $\displaystyle α (ABED)=40$, what is $\displaystyle α (ΔABC)$?
(5 marks)
12.    (a) The position vectors, relative to an origin $\displaystyle O$, of three nonlinear points $\displaystyle A, B$ and $\displaystyle C$ are $-2\hat{\text{i}}+3\hat{\text{j}}$, $3\hat{\text{i}}+2\hat{\text{j}}$ and $-\hat{\text{i}}-5\hat{\text{j}}$. Show that $\displaystyle ΔABC$ is isosceles.
(5 marks)
(b) The tangent at the point $\displaystyle C$ on the circle meets the diameter $\displaystyle AB$ produced at $\displaystyle T$. If $\displaystyle ∠BCT=27°$, calculate $\displaystyle ∠CTA$. If $\displaystyle CT=t$ and $\displaystyle BT=x$, prove that the radius of the circle is $\displaystyle {\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}}$.
(5 marks)
13.   (a) Find the angles of a triangle, given that $\displaystyle \angle A$ is obtuse and $\sec (B+C)=\operatorname{cosec}(B-C)=2$.
(5 marks)
(b) Differentiate $\displaystyle \frac{1}{\sqrt{x}}$ with respect to $\displaystyle x$ from the first principles.
(5 marks)
14.   (a) A cruise ship travels at a bearing of $\displaystyle 45°$ at $\displaystyle 15$ mph for $\displaystyle 3$ hours, and changes course to a bearing of $\displaystyle 120°$. It then travels $\displaystyle 10$ mph for $\displaystyle 2$ hours. Find the distance of the ship from its original position and also its bearing from the original position.
(5 marks)
(b) Find the two positive numbers $\displaystyle x$ and $\displaystyle y$ such that their sum is $\displaystyle 60$ and $\displaystyle xy^3$ is maximum.
(5 marks)
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ထူးချွန်စွာ အောင်မြင်ကြပါစေ

ေပးထားတဲ့ curve (အစိမ္းေရာင္) က $\displaystyle y=f(x)$ ျဖစ္ပါတယ္။ ခရမ္းေရာင္ $\displaystyle PQ$ မ်ဥ္းကေတာ့ $\displaystyle \text{curve}$ ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ခုကို ျဖတ္သြားလို႔ $\displaystyle \text{secant}$ လို႔ ေခၚမယ္။ $\displaystyle P$ အမွတ္မွာ $\displaystyle \text{curve}$ ကို ထိသြားတဲ့ အနက္ေရာင္မ်ဥ္းကိုေတာ့ $\displaystyle \text{tangent}$ လို႔ေခၚမယ္။

Given (ေပးခ်က္) : Curve : $\displaystyle y=f(x)$

Claim (ရွာရန္) : Gradient (Slope) of tangent at P

Explanation (ရွင္းလင္းခ်က္)

ေပးထားတာ curve : $\displaystyle y=f(x)$  ရွိတာေၾကာင့္ curve equation ထဲကို $\displaystyle x$ ေတြထည့္ရင္ $\displaystyle y$ ရမွာေပါ့။ တနည္းေျပာရင္ curve ေပၚမွာရွိတဲ့ အမွတ္ေတြကို ႀကိဳက္သေလာက္ ရွာႏိုင္ပါတယ္။ curve ေပၚမွာရွိတဲ့ အမွတ္ႏွစ္ကို ျဖတ္ဆြဲရင္ secant ေပါ့။ အမွတ္ႏွစ္မွတ္ သိမွေတာ့ gradient (slope) ကိုလည္း ရွာလို႔ရၿပီေပါ့။

$\displaystyle \begin{array}{l}\ \ \ y=f(x)\\\\\ \ \ {{y}_{1}}=f({{x}_{1}})\Rightarrow y+\delta y=f(x+\delta x)\\\\\therefore \ \text{Gradient of secant}\ PQ\\\\=\frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}\\\\=\frac{{y+\delta y-y}}{{x+\delta x-x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{x+\delta x-x}}\\\\=\frac{{\delta y}}{{\delta x}}\ \text{or }\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$

ဒါေပမယ့္ အခုလိုခ်င္တာက gradient of tangent ျဖစ္ပါတယ္။ secant မဟုတ္ဘူး။ ျပသနာက tangent က curve ေပၚမွာ အမွတ္တစ္မွတ္ကိုပဲ ထိသြားတာ၊ ႏွစ္မွတ္မရွိဘူး။ တစ္မွတ္ထဲကို ႏွစ္ကိုယ္ခြဲတြက္ေပါ့ လို႔ေျပာလိုက ေျပာႏိုင္ပါေသးတယ္။ ႏွစ္ကိုယ္ခြဲလိုက္မွေတာ့ $\displaystyle \frac{{{{y}_{1}}-y}}{{{{x}_{1}}-x}}=\frac{{y-y}}{{{{y}_{1}}-x}}=\frac{0}{0}$ ဆိုတာ indeterminate form ျဖစ္သြားၿပီ ဘာမွ ဆက္လုပ္လို႔မရေတာ့ဘူး။

ေရ တစ္စည္ထဲကို ေရတစ္ခြက္ ေပါင္းထည့္လို႔၊ ေရတစ္စည္ထဲက ေရတစ္ခြက္ ခပ္ထုတ္လိုက္လို႔ ေရ တစ္စည္ကို တိုးလားတယ္ ေလ်ာ့သြားတယ္လို႔ ေျပာေလ့မရွိၾကပါဘူး။ ဘာေၾကာင့္လဲ ဆိုေတာ့ $\displaystyle \frac{{\operatorname{ေရတစ္ခြက္}}}{{\operatorname{ေရတစ္စည္}}}\approx0$ ျဖစ္တာေၾကာင့္ပါ။ သခၤ်ာ႐ွဳေထာင့္က ၾကည့္ရင္ေတာ့ တစ္ခြက္တိုးတိုး တစ္စက္ တိုးတိုး အတိုး ရွိတာေပါ့့။

အလားတူပါပဲ တစ္မွတ္ထဲပဲ ရွိတဲ့ tangent ရဲ့ gradient ကို မရွာႏိုင္ေပမယ့္ အမွတ္ $\displaystyle P$ နားကို အလြန္နီးကပ္ေနတဲ့ အမွတ္တစ္ခုကို ယူလိုက္ရင္ေတာ့ အမွတ္ႏွစ္ခု ျဖစ္သြားလို႔ gradient ရွာႏိုင္ၿပီေပါ့။  tangent ေတာ့မဟုတ္ဘူး tangent နား အလြန္ကပ္ေနတဲ့ secant ရဲ့ gradient ေပါ့။ Calculus မွာေတာ့ $\displaystyle Q$ က $\displaystyle P$ အနားကို လံုေလာက္ေအာင္ နီးကပ္သြားရင္ Gradient of tangent = Gradient of Secant လို႔ သတ္မွတ္ပါတယ္။

ပံုမွာ ျမင္ေတြ႔ရတဲ့ အတိုင္းေပါ့။ $\displaystyle Q$ က Curve တေလွ်ာက္ $\displaystyle P$ အနားကို ကပ္သြားဖို႔ $\displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ႔သြားဖို႔လိုပါတယ္။ $\displaystyle {{x}_{1}}=x+\delta x$ ျဖစ္တာေၾကာင့္ $\displaystyle {{{x}_{1}}}$ ရဲ့ တန္ဖိုး ေလ်ာ့သြားဖို႔ ဆိုတာက $\displaystyle {\delta x}$ တန္ဖိုး ေလ်ာ့သြားမွ ျဖစ္မွာေပါ့။ ပံုမွာ $\displaystyle {\delta x}$ တန္ဖိုးသတ္မွတ္ထားတဲ့ slider ကို ဘယ္ဘက္ကို ေရႊ႕ၾကည့္ပါ။

$\displaystyle \begin{array}{*{20}{l}} {\text{When }\delta x\to 0,\ } \\ {} \\ {\text{Gradient of secant}\to \text{Gradient of tangent}} \\ {} \\ {\text{Therefore the gradient of secant approaches }} \\ {\text{the gradient of tangent when }\delta x\ \text{approaches 0}\text{.}} \\ {} \\ \begin{array}{l}\text{By limit notation,}\\\text{ }\end{array} \\ \begin{array}{l}\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\\\\text{Gradient of tangent =}\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array} \end{array}$

Gradient of tangent ကိုေတာ့ သေကၤတ $\displaystyle \frac{{dy}}{{dx}}$ (သို႔) $\displaystyle y'$ (သို႔) $\displaystyle f'(x)$ (သို႔) $\displaystyle \frac{d}{{dx}}\left[ {f(x)} \right]$ ျဖင့္သတ္မွတ္ပါတယ္။ ဒါ့ေၾကာင့္ ...

$\displaystyle \begin{array}{l}\frac{{dy}}{{dx}}={y}'=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}\\{f}'(x)=\frac{d}{{dx}}\left[ {f(x)} \right]=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}\end{array}$
If $\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$ and $\displaystyle a_{4}$ are the coefficients of any four consecutive terms in the expansion of $\displaystyle (1 + x)^n$, show that $\Large {\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}}$.

$\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$ , $\displaystyle a_{4}$ တို႔ဟာ $\displaystyle (1 + x)^n$ ျဖန္႔လိုက္တဲ့အခါ ရလာမယ့္ ကိန္းတန္းရဲ့ အစဥ္လိုက္ရွိေသာ ကိန္း ေလးလံုး (မည္သည့္ ဆက္တိုက္ ကိန္းေလးလံုးမဆို)   ေျမႇာက္ေဖၚကိန္းမ်ား ျဖစ္ၾကလွ်င္
$\displaystyle \large {\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}}$
ျဖစ္ေၾကာင္း သက္ေသျပပါ။

မည္သည့္ကိန္းေလးလံုးမဆို လို႔ ေျပာထားပါတယ္။ မထမကိန္း ​ေလးလံုးလို႔ မ​ေျပာပါဘူး။ မည့္သည့္ကိန္း ေလးလံုးမဆိုိ မွန္ကန္ရမွာ ျဖစ္ပါတယ္။ ဒါ့ေၾကာင့္တိက်တဲ့ ကိန္းေလးလံုးကို ယူၿပီး သက္ေသျပတာ ေမးခြန္းကေတာင္းဆိုခ်က္ ကို မျပည့္စံုေစပါဘူး။ မွားတယ္လို႔ ေျပာႏိုင္ပါတယ္။ general form နဲ႔ သက္ေသျပေပးရမွာ ျဖစ္ပါတယ္။
------------------------------------------------------------------------
Solution

$\displaystyle \begin{array}{l}{{(r+1)}^{{\text{th}}}}\text{term in the expansion of}\\{{(1+x)}^{n}}={}^{n}{{C}_{r}}{{x}^{r}}\ \\\\{{a}_{1}},{{a}_{2}},{{a}_{3}}\text{ and }{{a}_{4}}\ \text{are the coefficients }\\\text{of any four consecutive terms}\text{.}\\\\\therefore {{a}_{1}}={}^{n}{{C}_{r}},{{a}_{2}}={}^{n}{{C}_{{r+1}}},\\\ \ \ {{a}_{3}}={}^{n}{{C}_{{r+2}}}\ \text{and }{{a}_{4}}={}^{n}{{C}_{{r+3}}}\end{array}$

$\displaystyle \large{\begin{array}{l}\therefore \ \ \ \ \ \text{L}\text{.H}\text{.S}=\frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{}^{n}{{C}_{r}}}}{{{}^{n}{{C}_{r}}+{}^{n}{{C}_{{r+1}}}}}+\frac{{{}^{n}{{C}_{{r+2}}}}}{{{}^{n}{{C}_{{r+2}}}+{}^{n}{{C}_{{r+3}}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{}^{n}{{C}_{r}}}}{{{}^{n}{{C}_{r}}+{}^{n}{{C}_{r}}\frac{{n-(r+1)+1}}{{r+1}}}}+\frac{{{}^{n}{{C}_{{r+2}}}}}{{{}^{n}{{C}_{{r+2}}}+{}^{n}{{C}_{{r+2}}}\frac{{n-(r+3)+1}}{{r+1}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{^{n}{{C}_{r}}}}{{^{n}{{C}_{r}}\left( {1+\frac{{n-r}}{{r+1}}} \right)}}+\frac{{^{n}{{C}_{{r\text{ }+}}}_{2}}}{{^{n}{{C}_{{r\text{ }+}}}_{2}\left( {1+\frac{{n-r-2}}{{r+3}}} \right)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{{\frac{{r+1+n-r}}{{r+1}}}}+\frac{1}{{\frac{{r+3+n-r-2}}{{r+3}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{r+1}}{{n+1}}+\frac{{r+3}}{{n+1}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2(r+2)}}{{n+1}}\\\\\therefore \ \ \ \text{R}\text{.H}\text{.S}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}\ \ \ \\\ \ \\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}{{+}^{n}}{{C}_{{r\text{ }+}}}_{2}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}{{+}^{n}}{{C}_{{r\text{ }+}}}_{1}\frac{{n-(r+2)+1}}{{r+2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot {{\ }^{n}}{{C}_{{r\text{ }+}}}_{1}}}{{^{n}{{C}_{{r\text{ }+}}}_{1}\left( {1+\frac{{n-r-1}}{{r+2}}} \right)}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2\cdot \ }}{{\frac{{r+2+n-r-1}}{{r+2}}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\frac{{2(r+2)}}{{n+1}}\\\\\therefore \frac{{{{a}_{1}}}}{{{{a}_{1}}+{{a}_{2}}}}+\frac{{{{a}_{3}}}}{{{{a}_{3}}+{{a}_{4}}}}=\frac{{2{{a}_{2}}}}{{{{a}_{2}}+{{a}_{3}}}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array}}$
$\displaystyle \frac{{dy}}{{dx}}=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{\delta y}}{{\delta x}}$ or $\displaystyle {f}'(x)=\underset{{\delta x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$  ဘယ္လို ျဖစ္သြားလဲ...

A ship leaves harbour on a course N 72° E and after traveling for 50 metres, changes course to 108°. After a further 106 metres, find
(a) the distance of the ship from the harbour,
(b) its bearing from the harbour.

The cubic polynomial $\displaystyle f(x)$ is such that the coefficient of $\displaystyle x^3$ is $\displaystyle -1$ and the roots of the equation $\displaystyle f(x)=0$ are $\displaystyle 1$, $\displaystyle 2$ and $\displaystyle k$. Given that $\displaystyle f(x)$ has a remainder of $\displaystyle 8$ when divided by $\displaystyle x-3$, find the value of $\displaystyle k$ and the remainder when $\displaystyle f(x)$ is divided by $\displaystyle x+3$ .

$\displaystyle f(x)$ ဟာ အႀကီးဆံုးထပ္ကိန္း $\displaystyle 3$ ထပ္ပါတဲ့ $\displaystyle \text{polynomial}$ ကိန္းတန္း တစ္ခုျဖစ္တယ္။ $\displaystyle f(x)=0$ ကို ေျဖရွင္းလို႔ရတဲ့ $\displaystyle x$ တန္ဖိုးေတြက $\displaystyle 1$, $\displaystyle 2$ နဲ႔ $\displaystyle k$ ျဖစ္ၾကတယ္။ $\displaystyle x^3$ ရဲ့ ေျမႇာက္ေဖၚကိန္းက $\displaystyle -1$ ျဖစ္တယ္။

$\displaystyle f(x)$ ကို $\displaystyle x-3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $\displaystyle \text{(remainder)}$ က $\displaystyle 8$ ျဖစ္တယ္ဆိုရင္ $\displaystyle k$ ကို ရွာေပးပါ။

အဲဒီေနာက္ $\displaystyle f(x)$ ကို $\displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $\displaystyle \text{(remainder)}$ ကိုလည္း ရွာေပးပါ။

ဒီအခ်က္ေတြကိုေပါင္းလိုက္ရင္

• $\displaystyle f(x)=-1(x-1)(x-2)(x-k)$ လို႔သိရပါမယ္။

• $\displaystyle f(3)= 8$ ျဖစ္တယ္လို႔ သိရမယ္။

• $\displaystyle f(3)$ မွာ ကို $\displaystyle \text{function}$ မွာ အစားသြင္းၿပီး $\displaystyle f(3)$ နဲ႔ ညီေပးလိုက္ရင္ $\displaystyle k$ တန္ဖိုးရၿပီေပါ့။

• $\displaystyle k$ ရတဲ့အခါ $\displaystyle f(x)$ ကို $\displaystyle x+3$ နဲ႔ စားလို႔ရတဲ့ အႂကြင္း $\displaystyle \text{(remainder)}$ ဆိုတာ $\displaystyle f(-3)$ ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။
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Solution

$\displaystyle \begin{array}{l}\text{By the problem,}\\\\f(x)=-1(x-1)(x-2)(x-k)\\\\f(3)=8\\\\\therefore -1(3-1)(3-2)(3-k)=8\\\\\therefore -1(2)(1)(3-k)=8\\\\\therefore 3-k=-4\Rightarrow k=7\\\\\therefore f(x)=-1(x-1)(x-2)(x-7)\\\\\text{When }f(x)\ \text{is divided by }x+3,\\\\\text{the remainder = }f(-3)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-3-1)(-3-2)(-3-7)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-1(-4)(-5)(-10)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =200\end{array}$
Two circles intersect at $\displaystyle A$ and $\displaystyle B$. A point $\displaystyle P$ is taken on one so that $\displaystyle PA$ and $\displaystyle PB$ cut the other at $\displaystyle Q$ and $\displaystyle R$ respectively. The tangents at $\displaystyle Q$ and $\displaystyle R$ meet the tangent at $\displaystyle P$ in $\displaystyle S$ and $\displaystyle T$ respectively. Prove that
$\displaystyle \text{(a)}$ $\displaystyle \angle TPR=\angle BRQ$,
$\displaystyle \text{(b)}$ $\displaystyle PBQS$ is cyclic.

Two runners start from the same point at 12:00 noon, one of them heading north at 6 mph and the other heading 68° east of north at 8 mph. What is the distance between them at 3:00 that afternoon?

A ship is 5 km from a boat in a direction N 37o W and a lighthouse is 12 km from a boat in a direction S 53° W. Calculate the distance and direction of a ship from the lighthouse.

ဖန္တီးခဲ့တာေတာ့ၾကာပါၿပီ။ အသံုး၀င္လိမ့္မယ္ ထင္မိလို႔ ျပန္တင္ေပးလိုက္ပါတယ္။ Microsoft Power Point နဲ႔ ဖန္တီးၿပီး video အျဖစ္ save လိုက္လို႔ရပါတယ္။ Facebook ထက္ Youtube က resolution ပိုေကာင္းလို႔ Video File ေတြကို youtube မွာ တင္တာပိုေကာင္းပါတယ္။
Find the value of $\displaystyle m$ by matrix method for which the simultaneous equations $\displaystyle 3x + my = 5$ and $\displaystyle (m + 2) x + 5y = m$ have (i) an infinite number of solutions (ii) no solution.
Solution
$\displaystyle \left. \begin{array}{l}3x+my=5\\(m+2)x+5y=m\end{array} \right\}\ \ ---------(1)$

$\displaystyle \text{Transforming into matrix form},$

$\displaystyle \left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right)------(2)$

$\displaystyle \text{Let }A=\left( {\begin{array}{*{20}{c}} 3 & m \\ {m+2} & 5 \end{array}} \right),X=\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right)\ \text{and }B=\left( {\begin{array}{*{20}{c}} 5 \\ m \end{array}} \right).$

$\displaystyle \text{Then we have, }AX=B.$

$\displaystyle \text{If }\det A\ne 0,\text{then }{{A}^{{-1}}}\ \text{exists}\text{.}$

$\displaystyle {{A}^{{1}}}AX={{A}^{{-1}}}B$

$\displaystyle IX={{A}^{{-1}}}B$

$\displaystyle X={{A}^{{-1}}}B$ $\displaystyle \text{which is a unique solution for }X.$

$\displaystyle \begin{array}{l}\text{It is given that }X\text{ has}\\\text{(i) an infinite number of solutions and}\\\text{(ii) no solution}\text{.}\end{array}$

$\displaystyle \text{It means that the system has no unique solution}\text{.}$

$\displaystyle \text{Thus, }\det A=0.$

$\displaystyle \therefore 15-2m-{{m}^{2}}=0$

$\displaystyle \therefore (5+m)(3-m)=0$

$\displaystyle \therefore m=-5\ \text{or }m=3$

$\displaystyle \text{When }m=-5,\ \text{System (1) becomes}$

$\displaystyle 3x-5y=5\text{ and }-3x+5y=-5$

$\displaystyle \begin{array}{l}\therefore \text{The two equations represents the same }\\\text{straight line and there will be infinite }\\\text{number of solutions}\text{.}\end{array}$

$\displaystyle \text{When }m=3,\ \text{System (1) becomes}$

$\displaystyle 3x+3y=5\text{ and }5x+5y=-3$

$\displaystyle \begin{array}{l}\therefore \text{The two lines are parallel and }\\\text{there will be no solution}\text{.}\\\text{ }\end{array}$
 Find the equation of the normal to the curve $\displaystyle y = 1 - x^2$ at the point where the curve crosses the positive $\displaystyle x$-axis. Find also the coordinates of the point where the normal meets the curve again.

$\displaystyle y = 1 - x^2$ ဆိုတဲ့ curve က အေပါင္း $\displaystyle x$ ၀င္႐ိုးကို ျဖတ္သြားတဲ့ ေနရာမွာ ရွိတဲ့ normal line equation ကို ရွာေပးပါ။ ၎ normal line က curve ကို ေနာက္ တစ္ႀကိမ္ ျဖတ္သြားတဲ့ အမွတ္ကိုလည္း ရွာေပးပါ။

Curve က $\displaystyle x$ ၀င္႐ိုးကို $\displaystyle y=0$ ျဖစ္တဲ့ ေနရာမွာ ျဖတ္တာေပါ့။ ဒါဆိုရင္ ေပးထားတဲ့ curve : $\displaystyle 1 - x^2$ ကို 0 နဲ႔ ညီေပးလိုက္ရင္ curve ျဖတ္သြားတဲ့ $\displaystyle x$ ၀င္႐ိုးေပၚက အမွတ္ေတြ ရၿပီေပါ့။ အေပါင္း $\displaystyle x$ ၀င္႐ိုးလို႔ ေျပာထားတာလို႔ $\displaystyle x$ $\displaystyle \text{coordinate}$ မွာ အႏႈတ္ပါလာရင္ အႏႈတ္ တန္ဖိုးကို ပယ္ရပါမယ္။

Normal Line ရဲ့ equation ကို ရွာဖို႔ gradient (slope) ကိုလည္း သိရမယ္။ Normal Line ဆိုတာ tangent ေပၚ ေထာင့္မတ္က်လို႔ tangent gradient ရဲ့ အႏႈတ္လွန္ကိန္း (negative reciprocal) ျဖစ္ပါတယ္။ gradient of tangent =$\displaystyle m$ ျဖစ္တယ္ဆိုရင္ normal ရဲ့ gradient က $\displaystyle -\frac{1}{m}$ ျဖစ္မွာေပါ့။ ဒါဆိုရင္ tangent ရဲ့ gradient ကို သိရင္ normal line equation ကို $\displaystyle (y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})$ ဆိုတာနဲ႔ ရွာလိုက္ရင္ ရၿပီ။

Gradient of tangent = $\displaystyle \frac{dy}{dx}$ ဆိုတာ differentiation from the first principles ကတည္းက သိၿပီးျဖစ္မွာပါ။ tangent ရဲ့ gradient ကို ရဖို႔ ေပးထားတဲ့ curve ကို differentiate လုပ္ရင္ ရပါၿပီ။

Curve ကို differentiate လုပ္လို႕ရတဲ့ tangent ရဲ့ gradient ဆိုတာ general form (curve ေပၚမွာရွိတဲ့ မည္သည့္ေနရာကို မဆို ဆိုလိုတယ္)။ ေမးခြန္းက ေမးထားတာက အေပါင္း $\displaystyle x$ ၀င္႐ိုးကို ျဖတ္ေသာေနရာလို႔ ဆိုထားလို႔ ရွာထားတဲ့  $\displaystyle x$ ၀င္႐ိုး ျဖတ္မွတ္ [ $\displaystyle (x_{1},y_{1})$ လို႔ ဆိုၾကပါဆို႔] အစားသြင္းလိုက္မွ လိုခ်င္တဲ့ tangent line ရဲ့ gradient ကို ရမွာေပါ့။ ဆိုလိုတာက $\displaystyle m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}$ ပါ။

Normal Line equation ရရင္ ေမးခြန္းက ထပ္ဆင့္ေမးထားတာက ၎ Normal Line က Curve ကို ေနာက္တစ္ႀကိမ္ ျဖတ္တဲ့အမွတ္ကို ရွာေပးပါလို႔ ဆိုထားပါတယ္။

Line ႏွစ္ေၾကာင္းျဖတ္မွတ္ဆိုတာ အလယ္တန္း အဆင့္မွာကတည္းက သိခဲ့ၿပီးပါၿပီ။ ၎ Line ႏွစ္ေၾကာင္းကို တၿပိဳင္နက္ ေျပလည္ေစတဲ့ $\displaystyle x$ - $\displaystyle \text{coordinate}$ နဲ႔ $\displaystyle y$ - $\displaystyle \text{coordinate}$ ကို ရွာခိုင္းတာ ျဖစ္ပါတယ္။

ထိုနည္းတူ ပါပဲ။ Curve နဲ႔ Normal Line တို႔ျဖတ္မွတ္ဆိုတာ ၎ Equation ႏွစ္ေၾကာင္းကို တၿပိဳင္နက္ ေျပလည္ေစတဲ့ $\displaystyle x$ - $\displaystyle \text{coordinate}$ နဲ႔ $\displaystyle y$ - $\displaystyle \text{coordinate}$ ကို ရွာေပးရမွာ ျဖစ္ပါတယ္။ ဒါဆိုရင္ ေျဖရွင္းရမယ့္ အဆင့္ေတြကို သိေလာက္ၿပီလို႔ ထင္ပါတယ္။ တြက္ၾကည့္ၾကစို႔။

Solution

$\displaystyle \begin{array}{l}\text{Curve : }y=1-{{x}^{2}}\\\\\text{When the curve cuts the }x-\text{axis, }y=0.\\\\\therefore 1-{{x}^{2}}=0\Rightarrow {{x}^{2}}=1\Rightarrow x=\pm 1\\\\\text{But the required points lies on the }\\\text{positive }x-\text{axis, }\\\\\therefore x=1\\\\\text{Let }({{x}_{1}},{{y}_{1}})=(1,0)\\\\y=1-{{x}^{2}}\Rightarrow \frac{{dy}}{{dx}}=-2x\\\\\therefore m={{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{1}},{{y}_{1}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(1,0)}}}=-2(1)=-2\\\\\text{Hence the equation of normal line at }({{x}_{1}},{{y}_{1}})\ \text{is}\\\\(y-{{y}_{1}})=-\frac{1}{m}(x-{{x}_{1}})\\\\y-0=\frac{1}{2}(x-1)\Rightarrow y=\frac{1}{2}(x-1)\\\\\text{When this normal line meets the curve,}\\\\\frac{1}{2}(x-1)=1-{{x}^{2}}\Rightarrow 2{{x}^{2}}+x-3=0\\\\\therefore (2x+3)(x-1)=0\\\\\therefore x=-\frac{3}{2}\ \text{or}\ x=1\\\\\text{When }x=-\frac{3}{2}\ ,y=\frac{1}{2}(-\frac{3}{2}-1)=-\frac{5}{4}.\end{array}$

$\displaystyle \text{Therefore, the normal line meets}$
$\displaystyle \text{the curve again at the point (}-\frac{3}{2},-\frac{5}{4}).$
Find the equation of the tangent to the curve y=2x² which is parallel to the secant of the curve which passes through the points on the curve which have the coordinates x=−1 and x=2.

Solution
$\displaystyle \begin{array}{l}\text{Curve : }y=2{{x}^{2}}\\\\\text{When }x=-1,y=2{{(-1)}^{2}}=2\\\\\text{When }x=2,y=2{{(2)}^{2}}=8\\\\\therefore \text{Gradient of secant = }\frac{{8-2}}{{2-(-1)}}=2\\\\\text{The gradient of tangent is }\frac{{dy}}{{dx}}.\\\\\therefore \ \frac{{dy}}{{dx}}=4x\\\\\text{By the problem, tangent }\parallel \ \text{secant}\text{.}\\\\\therefore \ \frac{{dy}}{{dx}}=2\Rightarrow 4x=2\Rightarrow x=\frac{1}{2}\\\\\text{When }x=\frac{1}{2},y=2{{(\frac{1}{2})}^{2}}=\frac{1}{2}\end{array}$

$\displaystyle \begin{array}{l}\therefore \text{The tangent line touches the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\text{.}\\\\\therefore \text{Equation of tangent line to the curve at (}\frac{1}{2},\frac{1}{2}\text{)}\ \text{is}\\\\\ \ y-\ \frac{1}{2}=2(x-\frac{1}{2})\Rightarrow y=2x-\frac{1}{2}.\end{array}$
If a secant and a tangent aredrawn to a circle from an external point the square the of the tangent segment is equal to the product of the length of the secant segment and its external part. (Theorem-6 from grade 11 Mathematics TextBook)
ျပင္ပ အမွတ္တစ္ခုမွ စက္၀ိုင္းတစ္ခုသို႔ secant တစ္ေၾကာင္းႏွင့္ tangent တစ္ေၾကာင္း ဆြဲေသာအခါ secant ၏ တစ္ေၾကာင္းလံုးႏွင့္ အျပင္ဘက္ပိုင္း တု႔ိ၏ အလ်ားမ်ားေျမွာက္လဒ္သည္ tangent ၏အလ်ား ႏွစ္ထပ္ကိန္းႏွင့္ ညီသည္။

CE touches the circle BAED at E and circle CAB at C and DF touches the circle CAB at F. If CAD is a straight line, prove that CE² + DF² = CD².

$\displaystyle \begin{array}{l}\text{Proof : In smaller circle, }CAFB\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ C{{E}^{\text{2}}}=CA\cdot CD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{In larger circle, }BAED\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ D{{F}^{\text{2}}}=CD\cdot AD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore C{{E}^{\text{2}}}+D{{F}^{\text{2}}}=CA\cdot CD+CD\cdot AD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =CD\left( {CA+AD} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =CD\cdot CD\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore C{{E}^{\text{2}}}+D{{F}^{\text{2}}}=C{{D}^{\text{2}}}\end{array}$
Two circles intersect at $\displaystyle C$ and $\displaystyle D$. $\displaystyle ABCD$ is a cyclic quadrilateral in one circle. $\displaystyle BC$ produced meets the other circle at $\displaystyle E$. Te points $\displaystyle C, F, E$ and $\displaystyle D$ are concyclic. $\displaystyle AB$ produced meets $\displaystyle EF$ produced at $\displaystyle G$. Prove that $\displaystyle GFDA$ is a cyclic quadrilateral.
$\displaystyle \begin{array}{l}\text{Given }\ \text{ : }ABCD\text{ is a cyclic quadrilateral}\text{. }\\\text{ }\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ABG,BCE\ \text{and }EFG\ \text{are straight lines}\text{.}\\\\\text{To Prove : }GFDA\text{ is a cyclic quadrilateral}\text{.}\\\\\text{Proof : }\ \ \ ABCD\text{ is a cyclic quadrilateral}\text{.}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \theta +\beta =180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Since }D,C,F,E\ \text{are concyclic,}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta =\varepsilon \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{But }\varepsilon +\gamma =\beta \ \text{(Sum of interior }\angle \text{s of }\Delta \text{ = exterior }\angle \text{)}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta +\gamma =\beta \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \theta +\delta +\gamma =180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \angle ADF+\angle AGF=180{}^\circ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore D,C,F,E\ \text{are concyclic}.\end{array}$
The diagram shows triangle $\displaystyle OAB$ in which $\displaystyle \overrightarrow{{OA}}=\vec{a}$ and $\displaystyle \overrightarrow{{OB}}=\vec{b}$. The point $\displaystyle C$ is the midpoint of $\displaystyle OA$ and the point $\displaystyle D$ is the midpoint of $\displaystyle DC$.
(a) Express $\overrightarrow{{OD}}$ in terms of $\vec{a}$ and $\vec{b}$.
(b) If the point E lies on AB, then show that $\overrightarrow{{OE}}$ can be written in the form $\displaystyle \vec{a}+k\left( {\vec{b}-\vec{a}} \right)$, where k is a constant.
Given also that $\displaystyle OB$ produced meets $\displaystyle AB$ at $\displaystyle E$,
(c) find $\overrightarrow{{OE}}$,
(d) $\displaystyle AB:EB=2:1$
Solution

$\displaystyle \text{Since}$ $\displaystyle D$ $\displaystyle \text{is the midpoint of}$ $\displaystyle BC$, $\displaystyle \text{by midpoint formula,}$

$\displaystyle \begin{array}{l}\overrightarrow{{OD}}=\frac{1}{2}\left( {\overrightarrow{{OC}}+\overrightarrow{{OB}}} \right)\\\\\ \ \ \ \ \ =\frac{1}{2}\left( {\frac{1}{2}\vec{a}+\vec{b}} \right)\\\\\ \ \ \ \ \ =\frac{1}{4}\vec{a}+\frac{1}{2}\vec{b}\end{array}$

$\displaystyle \text{Let}$ $\displaystyle E$ $\displaystyle \text{be a point on}$ $\displaystyle AB$ $\displaystyle \text{such that}$ $\displaystyle \overrightarrow{{AE}}=k\overrightarrow{{AB}}.$

$\displaystyle \begin{array}{l}\therefore \overrightarrow{{OE}}=\overrightarrow{{OA}}+\overrightarrow{{AE}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\overrightarrow{{AB}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)\\\\\ \ \ \ \ \ \ \ =\vec{a}+k\left( {\vec{b}-\vec{a}} \right)\end{array}$

$\displaystyle \text{Since}$ $\displaystyle OD$ $\displaystyle \text{produced meets}$ $\displaystyle AB$ $\displaystyle \text{at}$ $\displaystyle E,$ $\displaystyle \text{let}$ $\displaystyle \overrightarrow{{OE}}=h\overrightarrow{{OD}}.$

$\displaystyle \therefore \left( {1-k} \right)\vec{a}+k\vec{b}=\frac{h}{4}\vec{a}+\frac{h}{2}\vec{b}$

$\displaystyle \text{Since}$ $\displaystyle \vec{a}$ $\displaystyle \text{and}$ $\displaystyle \vec{b}$ $\displaystyle \text{are not parallel and}$ $\displaystyle \vec{a}\ne \vec{0},\vec{b}\ne\vec{0},$

$\displaystyle \begin{array}{l}\ \ \ k=\frac{h}{2}\Rightarrow h=2k\\\\\ \ \ 1-k=\frac{h}{4}\Rightarrow h=4-4k\\\\\therefore 2k=4-4k\Rightarrow k=\frac{2}{3}\end{array}$

$\displaystyle \text{Since }\overrightarrow{{AE}}=k\overrightarrow{{AB}},$

$\displaystyle \begin{array}{l}\overrightarrow{{AE}}=\frac{2}{3}\overrightarrow{{AB}}\Rightarrow AE=\frac{2}{3}AB\\\\\therefore \overrightarrow{{EB}}=\frac{1}{3}\overrightarrow{{AB}}\Rightarrow EB=\frac{1}{3}AB\\\\\therefore AE:EB=\frac{2}{3}AB:\frac{1}{3}AB\\\\\therefore AE:EB=2:1\end{array}$
Two circles intersect at A and B. A line through A cuts the first circle at P and the second circle at Q. At P, a tangent PT is drawn and TQ produced meets the second circle again at R. Prove that the points P, T, R and B are concyclic.

စက္၀ိုင္းႏွစ္ခု $\displaystyle A$ နဲ႔ $\displaystyle B$ မွာ ျဖတ္ၾကပါတယ္။ $\displaystyle A$ ကို ျဖတ္ၿပီး မ်ဥ္းတစ္ေၾကာင္း ဆြဲရာမွာ ပထမစက္၀ိုင္းကို $\displaystyle P$ မွာ ျဖတ္ၿပီး ဒုတိယ စက္၀ိုင္းကို $\displaystyle Q$ မွာ ျဖတ္ပါတယ္။ အမွတ္ $\displaystyle P$ မွာ ၀န္းထိမ်ဥ္းတစ္ေၾကာင္း $\displaystyle PT$ ကို ဆြဲလိုက္ၿပီး တဖန္ အမွတ္ $\displaystyle T$ မွ $\displaystyle TQ$ ကို ဆက္ဆြဲရာ ဒုတိယ စက္၀ိုင္းကို $\displaystyle R$ မွာထပ္ၿပီး ျဖတ္ပါတယ္။ $\displaystyle P,T,R,B$  ဆိုတဲ့ အမွတ္ေလးခုဟာ စက္၀ိုင္းတစ္ခုထဲေပၚမွာ ရွိေၾကာင္း သက္ေသျပပါ။

 ဒီေမးခြန္းမွာ စက္၀ိုင္းႏွစ္ခု လို႔ပဲေျပာၿပီး စက္၀ိုင္းႏွစ္ခုဟာ အျခားကန္႔သတ္ခ်က္ မပါ၀င္ပါဘူး။ ဒါ့ေၾကာင့္ စက္၀ိုင္းႏွစ္ခုကို ဆြဲတဲ့ အခါ အရြယ္မတူ ထပ္တူမညီတဲ့ စက္၀ိုင္းႏွစ္ခု အျဖစ္ဆြဲသင့္ပါတယ္။ ထပ္တူမညီဘူး လို႔လည္း ေျပာမထားတဲ့ အတြက္ ထပ္တူညီတာ ဆြဲရင္ေကာ မရႏိုင္ဘူးလားလို႔ ေမးစရာ ရွိပါတယ္။ ဆြဲလို႔က ရႏိုင္ပါတယ္။ ဒါေပမယ့္ ထပ္တူညီတဲ့ ဂုဏ္သတၱိေတြကို သံုးခြင့္မရွိပါဘူး။ ထပ္တူညီျခင္းေၾကာင့္ (၁) အခ်င္း၀က္တူ၊ (၂) ထပ္တူညီေလးႀကိဳးမ်ား၊ (၃) ထပ္တူညီ အ၀န္းပိုင္းမ်ား ဆိုတဲ့ ဂုဏ္သတၱိေတြကို အျမင္အရ မွားယြင္း အသံုးျပဳမိတတ္ပါတယ္။ ေပးခ်က္အရ မပါ၀င္တဲ့ ဂုဏ္သတၱိ ေတြကို သံုးခြင့္မရွိပါဘူး။ ဒါေၾကာင့္ အပိုဂုဏ္သတၱိေတြ မပါ၀င္ေစတဲ့ ပံုမ်ိဳးကိုသာ ေရးဆြဲသင့္ပါတယ္။

ေမးခြန္းပါ ေပးခ်က္အရ ...

$\displaystyle PT$ က $\displaystyle \text{tangent}$ ျဖစ္ၿပီး $\displaystyle PA$ က ပထမစက္၀ိုင္းရဲ့ ေလးႀကိဳးျဖစ္တာေၾကာင့္ $\displaystyle \text{Theorem 4}$ ကို သံုးႏိုင္တဲ့ အခြင့္အေရး ရွိပါတယ္။

$\displaystyle PTQ$ က ႀတိဂံျဖစ္တာေၾကာင့္ အတြင္းေထာင့္မ်ား ေပါင္းျခင္း $\displaystyle 180{}^\circ$ ျဖစ္တယ္ ဆိုတဲ့ ဂုဏ္သတၱိ၊ ႀတိဂံ၏ အျပင္ေထာင့္သည္ အတြင္းမ်က္ႏွာခ်င္း ေထာင့္တစ္စံု ေပါင္းျခင္းနဲ႔ ညီတယ္ဆိုတဲ့ ဂုဏ္သတၱိမ်ားကို သံုးႏိုင္တဲ့ အခြင့္အေရး ရွိပါတယ္။

$\displaystyle P,T,R,B$ ကို $\displaystyle \text{concyclic}$ ျဖစ္ေၾကာင္း သက္ေသျပဖို႔ $\displaystyle P,T,R,B$ အမွတ္ ေလးမွတ္ကို ဆက္သြယ္ၿပီး စတုဂံ၏ အတြင္းေထာင့္ တစ္စံုေပါင္းျခင္း $\displaystyle 180{}^\circ$ ျဖစ္လွ်င္ ေထာင့္စြန္းမွတ္မ်ား စက္၀ိုင္းေပၚတြင္ က်ေရာက္သည္ (တစ္စက္၀န္းထဲ ရွိသည္) ဆိုတဲ့ $\displaystyle \text{Theorem 9}$ ကို သံုးသင့္တယ္လို႔ ခန္႔မွန္းတြက္ဆ သင့္ပါတယ္။ ပံုပါအခ်က္အလက္အရ $\displaystyle \text{Theorem 8}$ နဲ႔ $\displaystyle \text{Theorem 10}$ ကို သံုးဖို႔ အခြင့္အေရး နည္းတယ္လို႔ ခန္႔မွန္းႏိုင္ပါတယ္။

အဲဒီလိုတြက္ဆႏိုင္ရင္ သက္ေသျပဖို႔ လြယ္ကူသြားပါၿပီ။ သက္ေသျပၾကည့္ ရေအာင္။

$\displaystyle \text{Given :}\ \ PT\ \text{is a tangent and }PQR\ \text{is a secant}\text{.}$
$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ PAQ\text{ is a straight line}\text{.}$

$\displaystyle \text{To Prove : }P,T,R\ \text{and }B\ \text{are concyclic}\text{.}$

$\displaystyle \text{Proof : Draw }PB,AB\ \text{and }AR.\text{ }$

$\displaystyle \text{Since}PT\ \text{is a tangent and }PA\ \text{is a chord of first circle,}$

$\displaystyle {{\beta }_{1}}=\phi \ (\angle \ \text{between tangent and chord=}\angle \text{ in alt: segment})$

$\displaystyle \text{And }\theta =\phi +\delta (\ \text{Exterior }\angle \ \text{of }\Delta \text{ = Sum of opposite interior }\angle \text{s)}$

$\displaystyle \therefore \theta ={{\beta }_{1}}+\delta$

$\displaystyle \text{Since}ABRQ\ \text{is a cyclic quadrilateral,}$

$\displaystyle {{\beta }_{2}}+\theta =180{}^\circ$

$\displaystyle \therefore {{\beta }_{2}}+{{\beta }_{1}}+\delta =180{}^\circ \text{ }$

$\displaystyle \therefore \angle PBR+\angle PTR=180{}^\circ$

$\displaystyle \therefore P,T,R\ \text{and }B\ \text{are concyclic}\text{.}$

Exponential နဲ႔ Logarithmic Function ေတြကို differentiate လုပ္ရင္ ေအာက္ပါ Formula ေတြ သိထားဖို႔ လိုပါတယ္။

 $\displaystyle \begin{array}{*{20}{l}} \begin{array}{l}(1)\ \frac{d}{{dx}}({{a}^{x}})={{a}^{x}}\cdot \ln a,\ \\\ \ \ \ \ \operatorname{where}\ a>0\ \operatorname{and}\ a\ne 1.\end{array} \\ {} \\ {(2)\ \frac{d}{{dx}}({{e}^{x}})={{e}^{x}}} \\ {} \\ {(3)\ \frac{d}{{dx}}({{{\log }}_{b}}x)=\frac{1}{x}{{{\log }}_{b}}e} \\ {} \\ {(4)\ \frac{d}{{dx}}(\ln x)=\frac{1}{x}} \end{array}$

ဒါ့အျပင္ logarithm ရဲ့ basic rule တစ္ခ်ိဳ႕ျဖစ္တယ္ ေအာက္ပါ ဥပေဒသေတြကိုလည္း သိထားရပါမယ္။

 $\displaystyle \begin{array}{*{20}{l}} {(1)\ \ {{{\log }}_{b}}{{b}^{x}}=x} \\ {} \\ {(2)\ \ {{{\log }}_{b}}{{a}^{x}}=x{{{\log }}_{b}}a} \\ {} \\ \begin{array}{l}(3)\ \ {{\log }_{b}}(xy)={{\log }_{b}}x+{{\log }_{b}}y\\\\(4)\ \ {{\log }_{b}}(\frac{x}{y})={{\log }_{b}}x-{{\log }_{b}}y\end{array} \end{array}$

Question : If $\displaystyle y={{x}^{x}}$, prove that $\displaystyle \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}-\frac{y}{x}=0$.

Solution

$\displaystyle \begin{array}{l}\ \ \ y={{x}^{x}}\\\\\therefore \ln y=\ln {{x}^{x}}\\\\\therefore \ln y=x\ln x\end{array}$

$\displaystyle \operatorname{Differentiate \ with \ respect \ to \ x}$

$\displaystyle \ \ \frac{1}{y}\ \frac{{dy}}{{dx}}=x\frac{d}{{dx}}(\ln x)+\ln x\frac{d}{{dx}}(x)$

$\displaystyle \ \ \frac{1}{y}\ \frac{{dy}}{{dx}}=x\cdot \frac{1}{x}+\ln x(1)$

$\displaystyle \frac{1}{y}\ \frac{{dy}}{{dx}}=1+\ln x$

$\displaystyle \operatorname{Differentiate \ again \ with \ respect \ to \ x}$

$\displaystyle \ \ \frac{1}{y}\cdot \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)+\frac{{dy}}{{dx}}\cdot \frac{d}{{dx}}\left( {\frac{1}{y}} \right)=\frac{d}{{dx}}(1)+\frac{d}{{dx}}\left( {\ln x} \right)$

$\displaystyle \ \ \frac{1}{y}\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{{{{y}^{2}}}}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=0+\frac{1}{x}$

$\displaystyle \ \ \frac{1}{y}\frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{{{{y}^{2}}}}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=\frac{1}{x}$

$\displaystyle \operatorname{Multiplying \ both \ sides \ with \ y,}$

$\displaystyle \ \ \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}=\frac{y}{x}$

$\displaystyle \therefore \frac{{{{d}^{2}}y}}{{d{{x}^{2}}}}-\frac{1}{y}{{\left( {\frac{{dy}}{{dx}}} \right)}^{2}}-\frac{y}{x}=0$

အကယ္၍ $\displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$ ၊ $\displaystyle I=\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)$ နဲ႔ $\displaystyle xI-A$ ဟာ singular matrix ျဖစ္မယ္ဆိုရင္

$\displaystyle \begin{array}{l}xI-A=x\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {x-a} & b \\ c & {x-d} \end{array}} \right)\end{array}$

ဒါဆိုရင္ $\displaystyle xI-A$ ရဲ့ determinant ကို ရွာၾကည့္မယ္။

$\displaystyle \begin{array}{l} \det (xI-A)=(x-a)(x-d)-bc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{x}^{2}}-(a+d)x+ad-bc\end{array}$

ဒီေနရာမွာ $\displaystyle a+d$ ကို matrix  $\displaystyle A$ ရဲ့ trace လို႔ ေခၚၿပီး  $\displaystyle \operatorname{tr}(A)$ လို႔ သတ္မွတ္ပါမယ္။ $\displaystyle ad-bc$ ကေတာ့ matrix $\displaystyle A$ ရဲ့ determinant ျဖစ္ၿပီး $\displaystyle \det (A)$ လို႔ သတ္မွတ္ပါမယ္။

$\displaystyle xI-A$ ဟာ singular matrix ျဖစ္လို႔ $\displaystyle \det (xI-A)=0$ ျဖစ္မွာေပါ့။ ဒါဆိုရင္

$\displaystyle \begin{array}{l}\ \ \ \ \ \det (xI-A)=0\\\\\therefore {{x}^{2}}-(a+d)x+ad-bc=0\\\\\therefore {{x}^{2}}-\operatorname{tr}(A)x+\det (A)=0\end{array}$

အျမင္ရွင္းလင္း လြယ္ကူေအာင္ $\displaystyle \operatorname{tr}(A)=a+d$ ကို $p$ လို႔ထားၿပီး $\displaystyle \det (A)=ad-bc$ ကိုေတာ့ $q$ လို႔ထားလိုက္မယ္။

ဒါ့ေၾကာင့္ $\displaystyle {{x}^{2}}-px+q=0$ ဆိုတဲ့ polynomial equation တစ္ခု ရပါတယ္။ ၎ equation ကို characteristic equation လို႔ ေခၚပါတယ္။

Chareateristic polynomial ကို $\displaystyle f(x)$ လို႔ထားလိုက္မယ္ဆိုရင္ characteristic equation က $\displaystyle f(x)=0$ ေပါ့။ $\displaystyle x$ ေနရာမွာ matrix $\displaystyle A$ ကို အစားသြင္းလိုက္ရင္

$\displaystyle f(A)=0$
$\displaystyle \therefore {{A}^{2}}-pA+qI=O$

real number မွာေတာ့ အေျမႇာက္ထပ္တူရကိန္းက 1 ျဖစ္ေသာ္လည္း matrix မွာေတာ့ အေျမႇာက္ ထပ္တူရ matrix က identity matrix  $(I)$ ျဖစ္ တယ္ဆိုတာ သတိျပဳဖို႔ လိုပါတယ္။

ဒါ့ေၾကာင့္ မည္သည့္ square matrix မဆို ၎ရဲ့ characteristic equation ကို ေျပလည္ေစပါတယ္။

 $\displaystyle {{{A}^{2}}-\operatorname{tr}(A)A+\det (A)I=O}$

For 2× 2 matrix $\displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$,

 $\displaystyle {{A}^{2}}-(a+d)A+(ad-bc)I=O$

Proof : If $\displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$, then

$\displaystyle \begin{array}{l}\ \ \ {{A}^{2}}-(a+d)A+(ad-bc)I\\\\=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)-(a+d)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)+(ad-bc)\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+ad} & {ab+bd} \\ {ac+cd} & {ad+{{d}^{2}}} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {ad-bc} & 0 \\ 0 & {ad-bc} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$

Extension of Characteristic Equation
$\displaystyle \ \ \ \ {{A}^{2}}-pA+qI=O$

ႏွစ္ဘက္လံုးကို $\displaystyle {{A}^{{-1}}}$ နဲ႔ ေျမႇာက္လိုက္ရင္

$\displaystyle \begin{array}{l}\ \ \ {{A}^{2}}{{A}^{{-1}}}-pA{{A}^{{-1}}}+qI{{A}^{{-1}}}=O\\\\\ \ \ AA{{A}^{{-1}}}-pA{{A}^{{-1}}}+qI{{A}^{{-1}}}=O\\\\\ \ \ AI-pI+q{{A}^{{-1}}}=O\\\\\therefore A-pI+q{{A}^{{-1}}}=O\\\\\therefore A-(a+d)I+(ad-bc){{A}^{{-1}}}=O\end{array}$

အကယ္၍ $\displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)$ ျဖစ္ခဲ့ရင္ $\displaystyle {\operatorname{tr}(A)=p=a+d}$ and $\displaystyle \det (A)=q=ad-bc$ ဆိုတာ ေျပာခဲ့ၿပီးပါၿပီ။

$\displaystyle \begin{array}{l}\ \ \ {{A}^{2}}=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(1)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)\\\\\ \ {{A}^{{-1}}}=\frac{1}{{ad-bc}}\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\\\\\ \ pq{{A}^{{-1}}}=(a+d)(ad-bc)\frac{1}{{ad-bc}}\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\ \ \\\\\ \ \ \ \ \ \ \ \ \ \ =(a+d)\left( {\begin{array}{*{20}{c}} d & {-b} \\ {-c} & a \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {ad+{{d}^{2}}} & {-ab-bd} \\ {-ac-cd} & {{{a}^{2}}+ad} \end{array}} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -----(2)\\\\\ \ \ (q-{{p}^{2}})I=\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left[ {-bc-ad-{{a}^{2}}-{{d}^{2}}} \right]\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-bc-ad-{{a}^{2}}-{{d}^{2}}} & 0 \\ 0 & {-bc-ad-{{a}^{2}}-{{d}^{2}}} \end{array}} \right)--(3)\end{array}$

ညီမွ်ျခင္း (1), (2), (3) ကို ေပါင္းလိုက္ရင္
$\displaystyle \begin{array}{l}\ \ \ {{A}^{2}}+pq{{A}^{{-1}}}+(q-{{p}^{2}})I\\\\=\left( {\begin{array}{*{20}{c}} {{{a}^{2}}+bc} & {ab+bd} \\ {ac+cd} & {bc+{{d}^{2}}} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {ad+{{d}^{2}}} & {-ab-bd} \\ {-ac-cd} & {{{a}^{2}}+ad} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-bc-ad-{{a}^{2}}-{{d}^{2}}} & 0 \\ 0 & {-bc-ad-{{a}^{2}}-{{d}^{2}}} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$

$\displaystyle \begin{array}{l}\therefore {{A}^{2}}+pq{{A}^{{-1}}}+(q-{{p}^{2}})I=O\\\\\therefore {{A}^{2}}+(a+d)(ad-bc){{A}^{{-1}}}+\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I=O\end{array}$

အားလံုးျပန္ခ်ဳပ္လိုက္ရင္ $\displaystyle A=\left( {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right)\$ ျဖစ္ခဲ့ရင္

 $\displaystyle \begin{array}{l}(1){{A}^{2}}-(a+d)A+(ad-bc)I=O\\\\(2){A}-(a+d)I+(ad-bc){{A}^{{-1}}}=O\\\\(3){{A}^{2}}+(a+d)(ad-bc){{A}^{{-1}}}+\left[ {(ad-bc)-{{{(a+d)}}^{2}}} \right]I=O\end{array}$

Example : If $\displaystyle A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)\$ then

$\displaystyle \begin{array}{l}(1)\ \ {{A}^{2}}-(-2+4)A+(-8+9)I=O\\\\\Rightarrow {{A}^{2}}-2A+I=O\\\\(2){A}-(-2+4)I+(-8+9){{A}^{{-1}}}=O\\\\\Rightarrow {A}+{{A}^{{-1}}}-2I=O\\\\(3){{A}^{2}}+(-2+4)(-8+9){{A}^{{-1}}}+\left[ {(-8+9)-{{{(-2+4)}}^{2}}} \right]I=O\\\\\Rightarrow {{A}^{2}}+2{{A}^{{-1}}}-3I=O\end{array}$.

ေမးခြန္းျပန္လုပ္ေသာ္

$\displaystyle \begin{array}{l}\text{If}\ A=\left( {\begin{array}{*{20}{c}} {-2} & 3 \\ {-3} & 4 \end{array}} \right)\ \ \text{show that}\\\begin{array}{*{20}{l}} {(1){{A}^{2}}-2A+I=O} \\ {(2)A+{{A}^{{-1}}}-2I=O} \\ {(3){{A}^{2}}+2{{A}^{{-1}}}-3I=O} \end{array}\\\text{where}\ I\ \text{is a unit matrix of order 2}\text{.}\end{array}$

ဆရာႀကီး Dr. Shwe Kyaw ပို႔ခ်ခဲ့ေသာ post ကို မွီျငမ္း၍ ျပန္လည္ တင္ျပပါသည္။

Show that $\displaystyle A=\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)$ satisfies the equation $\displaystyle {{A}^{2}}-3A-7I=O$ where $\displaystyle I$ is a unit matrix of order 2. Hence using this result, find $\displaystyle {{A}^{{-1}}}$.

Solution

$\displaystyle \ \ \ \ A\ \ =\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)$

$\displaystyle \begin{array}{l}\therefore {{A}^{2}}=\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {25-3} \\ {-5+2} \end{array}\ \ \ \begin{array}{*{20}{c}} {15-6} \\ {-3+4} \end{array}} \right)\\\\\ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)\end{array}$

$\displaystyle \begin{array}{l}\therefore {{A}^{2}}-3A-7I\\\\=\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)-3\left( {\begin{array}{*{20}{c}} 5 & 3 \\ {-1} & 2 \end{array}} \right)-7\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {22} & 9 \\ {-3} & 1 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-15} \\ 3 \end{array}\ \ \ \begin{array}{*{20}{c}} {-9} \\ 6 \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-7} & 0 \\ 0 & {-7} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} {22-15-7} & {9-9} \\ {-3+3} & {1+6-7} \end{array}} \right)\\\\=\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 0 & 0 \end{array}} \right)\\\\=O\end{array}$

$\displaystyle \therefore {{A}^{2}}-3A-7I=O$

Multiplying both sides with $\displaystyle {{A}^{{-1}}}$,

$\displaystyle \begin{array}{l}{{A}^{2}}-3A-7I=O\\\\AA{{A}^{{-1}}}-3A{{A}^{{-1}}}-7I{{A}^{{-1}}}=O{{A}^{{-1}}}\\\\AI-3I-7{{A}^{{-1}}}=O\\\\AI-3I=7{{A}^{{-1}}}\end{array}$

$\displaystyle \begin{array}{l}\therefore 7{{A}^{{-1}}}=AI-3I\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-2} \end{array}} \right)-3\left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & 1 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {5-3} \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-2-3} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 2 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-5} \end{array}} \right)\\\\\therefore {{A}^{{-1}}}\ \ =\frac{1}{7}\left( {\begin{array}{*{20}{c}} 2 \\ {-1} \end{array}\ \ \ \begin{array}{*{20}{c}} 3 \\ {-5} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {\frac{2}{7}} & {\frac{3}{7}} \\ {-\frac{1}{7}} & {-\frac{5}{7}} \end{array}} \right)\end{array}$