Given a*b = LCM(a,b)
∴ (5*7) = LCM(5, 7) = 35

We have, y = a^x +x^a+x^x+a^a
Let v = x^x
log v = x log x
\(\frac { 1 }{ v } \frac { dv }{ dx } =x+\frac { 1 }{ x } +logx\)
\(\frac { dv }{ dx } =v\left| 1+logx \right| \)
= x^x(1+lodx)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

\(\left( \begin{matrix} 4 & -6 \\ 3 & -3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & -4 \\ 1 & -1 \end{matrix} \right)\)
Alternative Method :
We have \(\left( \begin{matrix} 4 & 2 \\ 3 & 3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & 0 \\ 1 & 1 \end{matrix} \right)\)
Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }+2{ C }_{ 1 }\)
\(\Rightarrow \ \left( \begin{matrix} 4 & -6 \\ 3 & -3 \end{matrix} \right) =\left( \begin{matrix} 1 & 2 \\ 0 & 3 \end{matrix} \right) \left( \begin{matrix} 2 & -4 \\ 1 & -1 \end{matrix} \right)\)

\({ tan }^{ -1 }\left[ \frac { \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } }\quad }{ \sqrt { 1+sin\quad x } +{ \sqrt { 1-sin\quad x } } } \right] \)
\(\begin{cases} \because 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 },0\le x\le \frac { \pi }{ 2 } \\ and\quad 1-sin\quad x={ \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } +\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } }{ \sqrt { { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } -\sqrt { { \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }^{ 2 } } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } +sin\frac { x }{ 2 } +cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } -cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left( \frac { 2cos\frac { x }{ 2 } }{ 2sin\frac { x }{ 2 } } \right) \)
\(={ tan }^{ -1 }\left( cot\frac { x }{ 2 } \right) \)
\(={ tan }^{ -1 }tan\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) =\left( \frac { \pi }{ 2 } -\frac { x }{ 2 } \right) \)

Reflexive Relation : A relation R on a set A is called reflexive relation if aRa for every \(a\in A\) ; if (a, a) \(\in \) R, for every \(a\in A\)
Example let
A = [1, 2, 3]
A x A =(1, 1) (1, 2)(1, 3) (2, 1) (2, 2) (2, 3) (3, 1)(3, 2) (3, 3)  \(\in R\) 
Since (a, a) \(\in R\) for every \(a\in A\)

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

The projection of \(\vec{a}\) on \(\vec{b} \vec{b}=\frac{\vec{a} \cdot \vec{b}}{\overrightarrow{|b|}}\) by the question \(\frac{(\lambda \hat{i}+\hat{j}+4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k})}{|2 \hat{i}+6 \hat{j}+3 \hat{k}|}=4\)
\( \Rightarrow \frac{(\lambda)(2)+(1)(6)+(4)(3)}{\sqrt{4+36+9}}=4 \)
\( \Rightarrow \frac{2 \lambda+6+12}{7}=4 \)
\( \Rightarrow 2 \lambda+18=28 \)
\( \Rightarrow 2 \lambda=10 .\)
Hence \(\lambda=5\)

\(={2\over3}x^{3/2}-{2\over5}x^{5/2}+c\)

Let  \(\int _{ 0 }^{ \pi /4 }{ (\sqrt { \tan { x } } +\sqrt { \cot { x } } ) } dx\)
\(=\int _{ 0 }^{ \pi /4 }{ \frac { \tan { x+1 } }{ \sqrt { \tan { x } } } } dx\)
\(Put \sqrt { \tan { x } } =t\ i.e. \tan { x } x={ t }^{ 2 }\)
so that \(\sec ^{ 2 }{ x } dx=2t\ dt \Rightarrow dx=\frac { 2t }{ 1+{ t }^{ 4 } } dt.\)
When \(x=0, t=0.\) When \(x=\frac { \pi }{ 4 } ,t=1.\)
\(\therefore I= \int _{ 0 }^{ 1 }{ \frac { { t }^{ 2 }+1 }{ t } } .\frac { 2t }{ 1+{ t }^{ 4 } } dt\)
\(=2\int _{ 0 }^{ 1 }{ \frac { { t }^{ 2 }+1 }{ { t }^{ 4 }+1 } } dx=2\int _{ 0 }^{ 1 }{ \frac { 1+1/{ t }^{ 2 } }{ { t }^{ 2 }+1/{ t }^{ 2 } } } dt\)
\(Put\ t-\frac { 1 }{ t } =y\) so that \(\left( 1+\frac { 1 }{ { t }^{ 2 } } \right) dt=dy.\)
Also \({ t }^{ 2 }-2+\frac { 1 }{ { t }^{ 2 } } ={ y }^{ 2 } \Rightarrow { t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } ={ y }^{ 2 }+2\)
\(\therefore I=2\int _{ t=0 }^{ 1 }{ \frac { dy }{ { y }^{ 2 }+2 } } \)
\(=\frac { 2 }{ \sqrt { 2 } } { \left[ \tan ^{ -1 }{ \frac { y }{ \sqrt { 2 } } } \right] }_{ t=0 }^{ 1 }\)
\(=\sqrt { 2 } { \left[ \tan ^{ -1 }{ \frac { 1 }{ \sqrt { 2 } } } \left( t-\frac { 1 }{ t } \right) \right] }_{ 0 }^{ 1 }\)
\(=\sqrt { 2 } \left[ \tan ^{ -1 }{ (0) } -\tan ^{ -1 }{ (-\infty ) } \right] \)
\(=\sqrt { 2 } \left[ \tan ^{ -1 }{ (-\infty ) } \right] =\sqrt { 2 } .\frac { \pi }{ 2 } .\)

(i) Let \(A= \left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right]\). Then \(A\prime =\left[ \begin{matrix} 5 & \frac { 1 }{ 2 } & -1 \end{matrix} \right] \).
(ii) Let \(A= \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\). Then \(A\prime =\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}\ \).
(iii) Let \(A= \left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt { 3 } & 5 & 6 \\ 2 & 3 & -1 \end{matrix} \right] \). Then \(A\prime =\left[ \begin{matrix} -1 & \sqrt { 3 } & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{matrix} \right] .\)

\(2\lambda +3=\pm 9\)

(i) P(both the students passing the examination)
\(=\frac{3}{5} \times \frac{4}{5}=\frac{12}{25}\)
(ii) P( only student A passing the examination)
\(=\frac{3}{5} \times \frac{1}{5}=\frac{3}{25}\)
(iii) P( only one of them passing the examination)
= p(A passes and B does not pass) or (A does not pass and B passes)
\(=\frac{3}{5} \times \frac{1}{5}+\frac{2}{5} \times \frac{4}{5}=\frac{3+8}{25}=\frac{11}{25}\)
(iv) P(none of them passing the examination)
= P (A does not pass and B does not pass)
\(=\frac{2}{5} \times \frac{1}{5}=\frac{2}{25}\)

\(f(x) =3 x^{4}-2 x^{3}-6 x^{2}+6 x+1 \)
\(f^{\prime}(x) =12 x^{3}-6 x^{2}-12 x+6 \)
\(=6\left(2 x^{3}-x^{2}-2 x+1\right) \)
\(=6\left(x^{2}-1\right)(2 x-1) \)
For a point of local maximum or minimum
\(f^{\prime}(x) =0 \)
\(\Rightarrow x =\pm 1, \frac{1}{2} \)
\(f^{\prime \prime}(x) =6\left[\left(x^{2}-1\right) 2+(2 x-1)(2 x)\right] \)
\(=6\left(5 x^{2}-2 x-2\right) \)
\(\text { when } x=1, f^{\prime \prime}(1)=6(5-2-2)>0\)
Therefore function attains local minimum at x = 1
Local minimum value
\(f(1)=3(1)^{4}-2(1)^{3}-6(1)^{2}+6(1)+1\)
\(=3-2-6+6+1=2\)
when x = -1 f'(-1) = 6(5 + 2 - 2) > 0
function attains local minimum at x = -1
Local minimum value = f(-1)
\(=3(-1)^{4}-2(-1)^{3}-6(-1)^{2}+6(-1)+1 \)
\(=3+2-6-6+1=-6 \)
\(\text { when } x =\frac{1}{2}, f^{\prime \prime}\left(\frac{1}{2}\right)=6\left(\frac{5}{4}-1-2\right)<0\)
\(\therefore \text { function attains local maximum at } x=\frac{1}{2}\)
\(\text { Local maximum value }=f\left(\frac{1}{2}\right)\)
\(=3\left(\frac{1}{2}\right)^{4}-2\left(\frac{1}{2}\right)^{3}-6\left(\frac{1}{2}\right)^{2}+6\left(\frac{1}{2}\right)+1 \)
\(=\frac{3}{16}-\frac{1}{4}-\frac{6}{4}+3+1 \)
\(=\frac{3}{16}-\frac{2}{4}+4=\frac{3-28+64}{15}=\frac{39}{16} \)

cos\((\frac {y}{x})\) = log |x| is the particular solution.

\(-5\overrightarrow { b } \)

Let 'x' and 'y' be the number of gold rings and chains respectively.
We have:
\(x\ge 0\)  ...(1) 
\(y\ge 0\)...(2)
\(x+y\le 24\)...(3) 
\(x+\frac { y }{ 2 } \le 16\) ...(4)
The objective function, or the profit, Z is:
Z = 300x + 190y ..(5)
We have to maximise Z subject to (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 24, 2x + y = 32.
The lines x + y = 24 and 2x + y = 32 meet at E (8,16).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Hence, the maximum profit is Rs. 5,440 and it is obtained when 8 gold rings and 16 chains are manufactured.

Eliminating y from the equations, we get

\(3 x-\frac{3}{2} x^{2}+12=0 \Rightarrow x^{2}-2 x-8=0 \)
\(\Rightarrow (x-4)(x+2)=0 \Rightarrow x=-2,4 \)
\(\text { Area }=\int_{-2}^{4}\left\{\frac{3 x+12}{2}-\frac{3}{4} x^{2}\right\} d x \)
\(=\left[\frac{3 x^{2}}{4}+6 x-\frac{x^{3}}{4}\right]_{-2}^{4} \)
\(=[(12+24-16)-(3-12+2)] \text { sq units } \)
\(=[20+7]=27 \text { sq units } \)

The given system of equations can be written as
AX = B
where A = \(\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \)
X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \)
\(\therefore\) X = A^-1B, if A^-1 exists
|A| = 40 => A^-1 exists
adj A = \(\left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
=> A^-1 = \(\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \)
\(\therefore\) X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
= \(\frac { 1 }{ 40 } \left[ \begin{matrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{matrix} \right] \left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right] \)
\(\therefore\) x = 1, y = 2 and z = -1

Let x and y be electronic and manually operated sewing machine purchased respectively.
\(\therefore\) LPP is Maximize.
P = 22x + 18y
Subject to, 
360x + 240y \(\ge \) 5,760
\(\Rightarrow\)  3x + 2y \(\le \) 48
x + y \(\le \) 20
x \(\ge \) 0 , y \(\ge \)  0

Vertices of feasible region are:
A (0, 20), B(8, 12), C(16, 0) & O(0, 0)
P(A) = 360, P(B) = 392. P(C) = 352
\(\therefore\) For Maximum P, Electronic machines = 8, and Manual machines = 12.

Given, \({ x }^{ 2 }dy+\left( xy+{ y }^{ 2 } \right) dx=0\)
\(\therefore \ \frac { dy }{ dx } =\frac { -\left( xy+{ y }^{ 2 } \right) }{ { x }^{ 2 } } \)
Put  y = vx
\(\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore \) The differential equation becomes
\(v+x\frac { dv }{ dx } =-\left( v+{ v }^{ 2 } \right) \)
\(\Rightarrow \frac { dv }{ { v }^{ 2 }+2v } =\frac { dx }{ x } \)
\(\Rightarrow \int { \frac { dv }{ \left( v+1 \right) ^{ 2 }-{ 1 }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
\(\Rightarrow \frac { 1 }{ 2 } \log { \frac { v }{ v+2 } } =-\log { x } +\log { C } \)
\(\Rightarrow \frac { C }{ x } =\sqrt { \frac { y }{ y+2x } } \)
If x = 1, y = 1 then \(c=\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow \frac { 1 }{ \sqrt { 3 } x } =\sqrt { \frac { y }{ y+2x } } \)

\(\int { \frac { \sqrt { x^{ 2 }+1 } \left\{ log({ x }^{ 2 }+1)-2logx \right\} }{ x^{ 4 } } } dx\)
\(=\sqrt { 1+\frac { 1 }{ { x }^{ 2 } } } \left( log\left( 1+\frac { 1 }{ x^{ 2 } } \right) \right) \frac { 1 }{ x^{ 3 } } dx\)
Let, \(1+\frac { 1 }{ { x }^{ 2 } } ={ t }^{ 2 }\)
\(\Rightarrow \frac { -2 }{ { x }^{ 3 } } dx=2tdt\)
\(\Rightarrow \frac { 1 }{ { x }^{ 3 } } dx=-tdt\)
\(=-\int { t(2logt)t\quad dt+ } -2\int { logt.{ t }^{ 2 }dt } \)
\(=-2logt.\frac { { t }^{ 3 } }{ 3 } +\int { 2\frac { 1 }{ t } .\frac { t^{ 3 } }{ 3 } dt } \)
\(=-\frac { 2 }{ 3 } logt.{ t }^{ 3 }+\frac { 2 }{ 9 } { t }^{ 3 }+C\)
\(=\frac { 2 }{ 3 } \left( 1+\frac { 1 }{ { x }^{ 2 } } \right) ^{ 3 }\left( -log\left( 1+\frac { 1 }{ { x }^{ 2 } } \right) +\frac { 1 }{ 3 } \right) +C\)

Equation of plane through the given lines is
\(\left\{ \vec { r } .\left( 2\hat { i } +2\hat { j } -3\hat { k } \right) -7 \right\} +\lambda \left\{ \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) -9 \right\} =0......\left( i \right) \)
\(\Rightarrow \vec { r } .\left\{ \left( 2+2\lambda \right) \hat { i } +\left( 2+5\lambda \right) \hat { j } +\left( -3+3\lambda \right) \hat { k } \right\} =\left( 7+9\lambda \right) .\left( ii \right) \)
Here, x intercept = z intercept 
\(\therefore \frac { 7+9\lambda }{ 2+2\lambda } =\frac { 7+9\lambda }{ -3+3\lambda } \Rightarrow \lambda =5\)
\(\therefore\)  Equation of plane in vector form is obtained by putting the value of in equation (ii).
i.e., \(\vec { r } .\left( 12\hat { i } +27\hat { j } +12\hat { k } \right) =52\) and equation of plane in cartesian form is
given as \(\left( x\hat { i } +y\hat { j } +z\hat { k } \right) .\left( 12\hat { i } +27\hat { j } +12\hat { k } \right) =52\)
i.e., 12x + 27y + 12z - 52 = 0

Let \(A=\left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] \)
A = AI
\(\therefore \left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & 3 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 1 }\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ 2C }_{ 1 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & -1 \\ 1 & 4 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 1 & -2 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ 2C }_{ 3 }\)
\({ C }_{ 2}\rightarrow { C }_{ 2 }+{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ -1 & 2 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ -3 & -3 & -2 \\ 2 & 2 & 1 \end{matrix} \right] \)
\({ C }_{ 3}\rightarrow { C }_{ 3 }+{ C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 1 \\ -3 & -3 & -5 \\ 2 & 2 & 3 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 }\)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)

\(|A|=\left[\begin{matrix}a&b\\c&{1+bc\over a}\end{matrix}\right]\)
\(=a\left(1+bc\over a\right)-bc =1+bc-bc=1\neq0\)
Since A is non-singular matrix ∴  A^-1 exists
Now \(A_{11}={1+bc\over a}; A_{12}=-b; A_{22}=a\)
\(\therefore\ adj.A=\begin{bmatrix}A _{11}&A_{12}\\A_{21}&A_{22} \end{bmatrix}'=\begin{bmatrix}{1+bc\over a}&-c\\-b&a \end{bmatrix}\)
\(=\begin{bmatrix} {1+bc\over a}&-b\\-c&a\end{bmatrix}.\)
\(\therefore\ A^{-1}={adj.A\over |A|}={1\over1}\begin{bmatrix} {1+bc\over a}&-b\\-c&a\end{bmatrix}\)
\(\therefore\ aA^{-1}=a\begin{bmatrix}{1+bc\over a}&-b\\-c&a \end{bmatrix}=\begin{bmatrix} 1+bc&-ab\\-ac&a^2\end{bmatrix}\)....(1)
And (a¹+bc+1)I-aA
\(=(a^2+bc+1)\begin{bmatrix} 1&0\\0&1\end{bmatrix}-a\begin{bmatrix} a&b\\c&{1+bc\over a}\end{bmatrix}\)
\(=\begin{bmatrix}a^2+bc+1&0\\0&a^2+bc+1 \end{bmatrix}=\begin{bmatrix}-a^2&-ab\\-ac&-(1+bc) \end{bmatrix}\)
\(=\begin{bmatrix}a^2+bc+1-a^2&0-ab\\0-ac&a^2+bc+-1-1bc \end{bmatrix}\)
\(=\begin{bmatrix}1+bc&-ab\\-ac&a^2 \end{bmatrix}\)..(2)
From (1) and (2), aA^-1 = (a² + bc + 1)I-aA

\(3A-5B=3\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] -5\left[ \begin{matrix} \frac { 2 }{ 3 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] -\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] .\)

\(A^{2}=\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 4 & 3 \\ 2 & 5 \end{array}\right]=\left[\begin{array}{cc} 16+6 & 12+15 \\ 8+10 & 6+25 \end{array}\right]=\left[\begin{array}{ll} 22 & 27 \\ 18 & 31 \end{array}\right]\)
\(A^{2}-x A+y I=O \)
\(\Rightarrow\left[\begin{array}{ll} 22 & 27 \\ 18 & 31 \end{array}\right]-\left[\begin{array}{ll} 4 x & 3 x \\ 2 x & 5 x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 0 & y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{ll} 22-4 x+y & 27-3 x+0 \\ 18-2 x+0 & 31-5 x+y \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\Rightarrow 27-3 x=0 \Rightarrow x=9 \)
\(\text {Also } 22-4 x+y=0 \)
y = 14