\(\left(x+\frac{1}{y}\right)^7 =7 C_0 x^7+7 C_1 x^6\left(\frac{1}{y}\right)+7 C_2 x^5\left(\frac{1}{y}\right)^2 +7 C_4 x^3\left(\frac{1}{y}\right)^4+7 C_5 x^2\left(\frac{1}{y}\right)^5 +7 C_6 x\left(\frac{1}{y}\right)^6+7 C_7\left(\frac{1}{y}\right)^7\)
\(=x^7+\frac{7 x^6}{y}+\frac{21 x^5}{y^2}+\frac{35 x^4}{y^3} +\frac{35 x^3}{y^4}+\frac{21 x^2}{y^5}+\frac{7 x}{y^6}+\frac{1}{y^7}\)

The system of equations can be written in the form AX = B where,
\(A=\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
Now, |A| = \(\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right| =1\left| \begin{matrix} 0 & 3 \\ -3 & 0 \end{matrix} \right| -2\left| \begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix} \right| +1 \left| \begin{matrix} 1 & 0 \\ 2 & -3 \end{matrix} \right| \)
= 1(0 + 9) - 2(0 - 6) + 1(-3 - 0) = 9 + 12 - 3 = 18 \(\neq \) 0 
\(\Rightarrow\) A^-1 exists.
A₁₁ = 0 + 9 = 9, A₁₂ = -(0 - 6) = 6, A₁₃ = -3 - 0 = -3
A₂₁ = -(0 + 3) = -3, A₂₂ = 0 - 2 = -2, A₂₃ = -(-3 - 4) = 7
A₃₁ = 6 - 0 = 6, A₃₂ = -(3 - 1) = -2, A₃₃ = 0 - 2 = -2
\(\therefore adj\quad A={ \left[ \begin{matrix} 9 & 6 & -3 \\ -3 & -2 & 7 \\ 6 & -2 & -2 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\(\therefore \ X={ A }^{ -1 }B=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 18 } \left[ \begin{matrix} 63 & -33 & +6 \\ 42 & -22 & -2 \\ -21 & +77 & -2 \end{matrix} \right] =\frac { 1 }{ 18 } \left[ \begin{matrix} 36 \\ 18 \\ 54 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore\) x = 2, y = 1, and z = 3.

To draw a line we need two points
No. of chords \(=21 C_2=\frac{21 \times 20}{2 \times 1}=210\)

A hexagon has 6 vertices and to draw a triangle we need 3 points
Number of triangles \(=6 C_3=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20\)

x² + y² = 16
\(\therefore\)  Centre is (0,0), r² = 16
r = 4 units

5x² + 5y² +4x - 8y - 16 = 0
[Divide by 5]
x² + y² +  \(\frac { 4 }{ 5 } x-\frac { 8 }{ 5 } y-\frac { 16 }{ 5 } =0\)
Here 2g  = \(\frac { 4 }{ 5 } \) \(\Rightarrow\)  \(g=+\frac { 2 }{ 5 } \)
2f = \(-\frac { 8 }{ 5 } \)     \(\Rightarrow\) \(f=-\frac { 4 }{ 5 } \)
and c = \(-\frac { 16 }{ 5 } \)
Center of the circle is (-g, -f) \(\Rightarrow \) \(\left( -\frac { 2 }{ 5 } ,\frac { 4 }{ 5 } \right) \)
Radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
\(\Rightarrow\) r = \(\sqrt { \frac { 4 }{ 25 } +\frac { 16 }{ 25 } +\frac { 16 }{ 5 } } =\sqrt { \frac { 20 }{ 25 }+ { \frac { 16 }{ 5 }} } \)
\(\Rightarrow\) \(r=\sqrt { \frac { 20 }{ 5 } } \) = \(\sqrt { 4 } \) = 2 units

(x + 2) ( x - 5) + (y -2 ) ( y -1) = 0
\(\Rightarrow\) x² -5x + 2x - 10 + y² - y - 2y + 2 = 0
\(\Rightarrow\)  x² + y² - 3x - 3y - 8 = 0
here 2g  = -3    \(\Rightarrow\)  \(g=-\frac { 3 }{ 2 }\)
2f  = -3   \(\Rightarrow\)   \(f=-\frac { 3 }{ 2 } \)
and C = -8
Center of the circle (-g, -f) = \(\left( \frac { 3 }{ 2 } ,\frac { 3 }{ 2 } \right) \)
A radius of the circle is \(\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } \)
Radius of the circle is  \(= \sqrt { \frac { 9 }{ 4 } +\frac { 9 }{ 4 } +8 }\)
\(=\sqrt{\frac{18}{4}+8}=\sqrt{\frac{9}{2}+8}\)
\(r=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}} \text { units }\)

Let \(\triangle=\begin{vmatrix} 6&-1&5\\3&0&4\\-2&7&-3 \end{vmatrix}\)
Now M₁₂ = \(\begin{vmatrix} 3&4\\-2&-3 \end{vmatrix}=-9-(-8)=-1\)
M₂₂ = \(\begin{vmatrix} 6&5\\-2&-3 \end{vmatrix}=-18-(-10)=-8\)
M₃₂ = \(\begin{vmatrix} 6&5\\3&4 \end{vmatrix}=24-15=9\)
Now expansion of |A| using co-factors of elements of second column we get,
|A| = a₁₂ A₁₂ + a₂₁ A₂₁ + a₃₁A₃₁
A₁₂ = (-1)^{1 + 2} M₁₂ (-1) (-1) = 1
A₂₂ = (-1)^{2 + 2} M₂₂ = 1(-8) = -8
A₃₂ = (-1)^{3 + 2} M₃₂ = -(9) = -9
\(\therefore\) |A| = -1(1) + 0(-8) + 7(-9) = -1 - 63 = -64.

In "AGAIN" the letters in ascending order are A, A, G, I, N
∴  The first word is AAGIN
Number of words beginning with A A
= No of ways of arranging G, 1, N = 3! = 6
The next word begin with AG and it is AGAIN
∴  No. of words before AGAIN = 6.
∴  Rank of word AGAIN = 7

No. of ways of giving I prize = n
No. of ways of giving II prize = n
.............................................
Similarly no. of ways of giving n^th prize = n
Therefore By fundamental principle of counting, the required number of ways 

The given lines are
2x - 5y + 3 = 0
5x - 9y + \(\lambda \) = 0
x - 2y + 1 = 0
The condition for the lines to be concurrent is
\(\left| \begin{matrix} 2 & -5 & 3 \\ 6 & -9 & \lambda \\ 1 & -2 & 1 \end{matrix} \right| =0\quad \)
Expanding along R₁ , we get
\(2\left| \begin{matrix} -9 & \lambda \\ -2 & 1 \end{matrix} \right| +5\left| \begin{matrix} 5 & \lambda \\ 1 & 1 \end{matrix} \right| +3\left| \begin{matrix} 5 & -9 \\ 1 & -2 \end{matrix} \right| =0\)
\(\Rightarrow 2(-9+2\lambda )+5(5-\lambda )+3(-10+9)=0\)
\(\Rightarrow -18+4\lambda +25-5\lambda -30+27=0\)
\(\Rightarrow -\lambda +4=0\Rightarrow \lambda =+4\)

The two lines are x (2 + 6k) + y(3 - k) + 4 + 12k = 0
7x + 5y - 4 = 0
Let m₁ and m₂ be the slopes of the given lines.
Then m₁=\(-\frac { { \text {Co-efficient of y} } }{ \text {Co-efficient of y }} =\frac { -(2+6k) }{ 3-k } \)
\(\\ and\ { m }_{ 2 }=\frac { -7 }{ 5 } \)
Since the given lines are perpendicular,
\(-\left( \frac { 2+6k }{ 3-k } \right) \left( \frac { -7 }{ 5 } \right) =-1\)
\(\Rightarrow 7(2+6k)=-5(3-k)\)
\(\Rightarrow 14+42k=-15+5k\)
\(\Rightarrow 37k=-29\)
\(\Rightarrow k=\frac { -29 }{ 37 } \)

Solution: Let p (x₁,y₁) be any point on the locus, such that the square of its distance from A (3, -2) is equal to its distance from 5x - 12y = 13.
\(\therefore\) (x₁ - 3)² + (y₁ + 2)² = \(\frac { \left| { 5x }_{ 1 }-12{ y }_{ 1 }+13 \right| }{ \sqrt { { 5 }^{ 2 }+{ \left( -12 \right) }^{ 2 } } } \)
⇒ 13[(x₁ - 3)² + (y₁ + 2)²] = 土(5x₁ - 12y₁ + 13)
⇒ 13(x₁² - 6x₁ + 9 + y₁² + 4 + 4y₁) = 士(5x₁ - 12y₁ + 13)
Case (i)
⇒ 13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = 5x₁ - 12y₁ + 13
⇒ 13x₁² + 13y₁² - 83x₁ + 64y₁ + 182 = 0
Case (ii)
13(x₁² + y₁² - 6x₁ + 4y₁ + 13) = -(5x₁ - 12y₁ + 13)
13x₁² + 13y₁² - 73x₁ + 40y₁ + 156 = 0
\(\therefore\) Locus of (x₁, y₁) is 13x² + 13y² - 83x + 64y + 182 = 0 (or) 13x² + 13y² - 73x + 40y + 156 = 0

Let P(x₁, y₁) be any point on the locus and let A(0, 2) B(0, -2) be the fixed points.
By the given condition, PA + PB =6
\(\Rightarrow \sqrt { { \left( { x }_{ 1 }-0 \right) }^{ 2 }+{ \left( { y }_{ 1 }-2 \right) }^{ 2 } } +\sqrt { { ({ x }_{ 1 }-0 })^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } =6\)
\(\Rightarrow \sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 } } =6-\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring both sides we get,
\({ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }-2) }^{ 2 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }=36-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } +{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }\)
\(\Rightarrow \quad { x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4-4{ y }_{ 1 }-36-{ x }_{ 1 }^{ 2 }-{ y }_{ 1 }^{ 2 }-4-4{ y }_{ 1 }=-12\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= -8y₁ - 36 = -12 \(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
= 2y₁ + 9 = 3\(\sqrt { { x }_{ 1 }^{ 2 }+{ ({ y }_{ 1 }+2) }^{ 2 } } \)
Squaring again we get,
(2y₁+9)² = 9[\({ x }_{ 1 }^{ 2 }+({ y }_{ 1 }+2)^{ 2 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }=9[{ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+4+4{ y }_{ 1 }]\)
\(\Rightarrow 4{ y }_{ 1 }^{ 2 }+81+36{ y }_{ 1 }-9{ x }_{ 1 }^{ 2 }-9{ y }_{ 1 }^{ 2 }-36-36{ y }_{ 1 }=0\)
\(\Rightarrow -9{ x }_{ 1 }^{ 2 }-5{ y }_{ 1 }^{ 2 }+45=0\)
\(\Rightarrow 9{ x }_{ 1 }^{ 2 }+5{ y }_{ 1 }^{ 2 }=45\)
\(\therefore \)  Locus of (x₁ , y₁) is 9x² + 5y² = 45

Slope of the line 3x-y+5=0 is
\(\Rightarrow \quad { m }_{ 2 }=-\frac { co-efficient\quad of\quad x }{ co-efficient\quad of\quad y } =\frac { -3 }{ -1 } =3\)
Let m₁ = m and \(\theta =45\)
\(\therefore \quad tan\theta =\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| \)
\(\Rightarrow \quad tan\quad 45=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1=\left| \frac { m-3 }{ 1+3m } \right| \)
\(\Rightarrow \quad 1|1+3m|=|m-3|\)
\(\Rightarrow \quad 1+3m=\pm (m-3)\)
\(\Rightarrow \quad 1+3m=m-3\quad or\quad 1+3m=-m+3\)
\(\Rightarrow\)   2m = - 4 or  4m = 2
\(\Rightarrow \)  m = - 2  or  \(m=\frac { 2 }{ 4 } =\frac { 1 }{ 2 }​​​​​​​\)
\(\therefore \)   m =  - 2 or \(\frac { 1 }{ 2 } \)

\(\text {Co-factor of }-7=\left|\begin{array}{cc} 2 & -3 \\ 6 & 0 \end{array}\right|=0+18=18\)

(Corresponding co-factor)

x + y = n
3 + 2 = 5 = n
nC₄ = 5C₄ = 5C₁ = 5

nP₂ = 20
n(n - 1) = 5 x 4
n = 5

Given A \(=\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=(4+7)-1(3-7)+1(-3-4)\)
\(=11+4-7=8\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 11 & 4 & -7 \\ -2 & 0 & +2 \\ 3 & -4 & 1 \end{array}\right)\)
\(\operatorname{adj} A=\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(\mathrm{A}(\operatorname{adj} A)=\left(\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right)\left(\begin{array}{ccc} 11 & -2 & 3 \\ 4 & 0 & -4 \\ -7 & 2 & 1 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 11+4-7 & -2+0+2 & 3-4+1 \\ 33+16-49 & -6+0+14 & 9-16+7 \\ 11-4-7 & -2+0+2 & 3+4+1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{array}\right)=8\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I_3\)
\((adj\ A)\ A=\begin{bmatrix}11&-2&3\\4&0&-4\\-7&2&1\end{bmatrix}\begin{bmatrix} 1&1&1\\3&4&7\\1&-1&1 \end{bmatrix}\)
\(=\begin{bmatrix} 11-6+3&11-8-3&11-14+3\\4+0-4&4+0+4&4+0-4\\-7+6+1&-7+8-1&-7+14+1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8 \end{bmatrix}\) ...(2)
\(|A|.{I}_{3}=8\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}=\begin{bmatrix} 8&0&0\\0&8&0\\0&0&8\end{bmatrix}\)  ...(3)
From (1), (2) and (3)
A (adj A) = (adj A) A = |A|I₃.

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

Compare the equation
12x² - 10xy + 2y² + 14x - 5y + 2 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We get a = 12, 2h = -10, b = 2, 2g = 14, 2f = -5
\(h=-5\quad g=7\quad f=-\frac { 5 }{ 2 } ,c=2\)
\(\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=\left|\begin{array}{ccc} 12 & -5 & 7 \\ -5 & 2 & \frac{-5}{2} \\ 7 & \frac{-5}{2} & 2 \end{array}\right|\)
\(=12\left(4-\frac{25}{4}\right)+5\left(-10+\frac{35}{2}\right)+7\left(\frac{25}{2}-14\right)\)
\(=48-75-50+\frac{175}{2}+\frac{175}{2}-98\)
= -175 + 175 = 0
Hence the given equations represent a pair of straight lines.
To find separate equation
\(12 x^2-10 x y+2 y^2 =12 x^2-6 x y-4 x y+2 y^2 \)
\(=6 x(2 x-y)-2 y(2 x-y) \)
\(=(6 x-2 y)(2 x-y)\)
\(12 x^2-10 x y+2 y^2+ 14 x-5 y+2 =(6 x-2 y+l)(2 x-y+\mathrm{m})\)
Comparing the coefficient of x and y
14 = 6m + 2l
divided by 2
7 = 3m + l .........(1)
-5 = -2m - l .......(2)
Solving (1) and (2) we get m = 2, 1 = 1
The separate equations are
6x - 2y + 1 = 0
2x - y + 2 = 0

\(\frac{5 x^2-8 x+5}{(x-2)\left(x^2-x+1\right)}=\frac{A}{x-2}+\frac{B x+C}{x^2-x+1}\)
\(\frac{5 x^2-8 x+5}{(x-2)\left(x^2-x+1\right)} =\frac{A\left(x^2-x+1\right)+(B x+C)(x-2)}{(x-2)\left(x^2-x+1\right)}\)
\(\Rightarrow\) 5x² - 8x + 5 = A (x² - x + 1) + (Bx + C) (x - 2)  ...(1)
Substituting x = 2 in (1) we get,
20 - 16 + 5 = A (4 - 2 + 1)
\(\Rightarrow\)  9 = A ( 3 )\(\Rightarrow\)  \(A={{9}\over3{}}\) \(\Rightarrow\)  \({A=3}\)
Equating the co-efficient of x² in (1) we get,
5 = A + B
\(5=3+B \Rightarrow B=2\)
\(\text {If } x=0\)
\(5=A-2 C\)
\(\Rightarrow\) 5 = 3 - 2C 
\(\Rightarrow\)  2 = - 2C
\(\Rightarrow\) c = -1
\(\therefore\) \({{5x^2-8x+5}\over{(x-2)(x^2-x+1)}}={{3}\over{x-2}}+{{2x-1}\over{x^2-x+1}}\)

To show that A is inverse of A itself it is enough if, we prove that
\(A . A=I \left(\because \mathrm{AA}^{-1}=\mathrm{I}\right. and \left.\mathrm{A}^{-1}=\mathrm{A}\right)\)
\(A \cdot A=\left(\begin{array}{ccc} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{array}\right)\left(\begin{array}{ccc} -1 & 2 & -2 \\ 4 & -3 & 4 \\ 4 & -4 & 5 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 1+8-8 & -2-6+8 & 2+8-10 \\ -4-12+16 & 8+9-16 & -8-12+20 \\ -4-16+20 & 8+12-20 & -8-16+25 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\therefore \mathrm{A} \text { is inverse of } \mathrm{A}\)

Given matrix A is singular, if |A| = 0
\(|A|=\begin{vmatrix} x+1&-3&4\\-5&x+2&2\\4&1&x-6 \end{vmatrix}=0\)
Expanding along R₁ we get,
\(|A|=x+1\begin{vmatrix}x+2 & 2 \\ 1 & x-6 \end{vmatrix}+3\begin{vmatrix} -5 & 2 \\ 4 & x-6 \end{vmatrix}+4\begin{vmatrix}-5 & x+2 \\ 4 & 1\end{vmatrix}=0\)
\(\Rightarrow\) (x + 1)[(x + 2)(x - 6) - 2] + 3[-5 (x - 6) - 8] + 4 [-5 - 4 (x + 2)] = 0
\(\Rightarrow\) (x + 1) [x² - 4x - 12 - 2] + 3[-5x + 30 - 8] + 4 [-5 - 4x - 8] = -0
\(\Rightarrow\) (x + 1)(x² - 4x - 14) + 3(-5x + 22) + 4(-4x - 13) = 0
\(\Rightarrow\) x³ - 4xl - 14x + xl - 4x - 14 - 15x + 66 - 16x - 52 = 0
\(\Rightarrow\) x³ - 3x² - 49x = 0
\(\Rightarrow\) x(x² - 3x - 49) = 0
\(\Rightarrow\)    \(x=0\ or\ x={{3\pm\sqrt{{(-3)}^{2}}-4(1)(-49)}\over{2a}}\)                                                           
\(\begin{bmatrix} \because\ x = {-b \pm \sqrt{b^2-4ac} \over 2a},a = 1, b = -3, c=-49 \end{bmatrix}\)
\(\Rightarrow\) \(x=0\ or\ x={{3\pm\sqrt{9+196}}\over{2}}\)
\(\Rightarrow\) \(x=0\ or\ x={{3\pm\sqrt{205}}\over{2}}\)

\(3 x-y+2 z=13 ; 2 x+y-z=3\)
\(x+3 y-5 z=-8\)
The given system can be written as
\(\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 3 & -1 & 2 \\ 2 & 1 & -1 \\ 1 & 3 & -5 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(|A|=3(-5+3)+1(-10+1)+2(6-1)\)
\(=-6-9+10=-5 \neq 0\)
\(\therefore A^{-1} exists\)
\(\mathrm{A}_{11}=\text {Co-factor of } 3=(-5+3)=-2 \)
\(\mathrm{A}_{12}=\text {Co-factor of }-1=-(-10+1)=9 \)
\(\mathrm{A}_{13}=\text {Co-factor of } 2=(6-1)=5\)
\(\mathrm{A}_{21}=\text { Co-factor of } 2=-(5-6)=1\)
\(\mathrm{A}_{22}=\text { Co-factor of } 1=-15-2=-17 \)
\(\mathrm{A}_{23}=\text { Co-factor of }-1=-(9+1)=-10\)
\(\mathrm{A}_{31}=\text {Co-factor of } 1=1-2=-1\)
\(\mathrm{A}_{32}=\text {Co-factor of } 3=-(-3-4)=7 \)
\(\mathrm{A}_{33}=\text {Co-factor of }-5=3+2=5\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 9 & 5 \\ 1 & -17 & -10 \\ -1 & 7 & 5 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\frac{1}{-5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\)
\(X=A^{-1} B=\frac{-1}{5}\left(\begin{array}{ccc} -2 & 1 & -1 \\ 9 & -17 & 7 \\ 5 & -10 & 5 \end{array}\right)\left(\begin{array}{c} 13 \\ 3 \\ -8 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{ccc} -26 & 3 & 8 \\ 117 & -51 & -56 \\ 65 & -30 & -40 \end{array}\right)\)
\(=\frac{-1}{5}\left(\begin{array}{c} -15 \\ 10 \\ -5 \end{array}\right)\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right)\)
\(x=3, \mathrm{y}=-2, \mathrm{z}=1\)

(y - 2)² = 4(x - 1).
\(\Rightarrow\) y² = 4x where X = x - 1 and Y = y - 2. \(\Rightarrow\) 4a = 4 \(\Rightarrow\) a = 1

Given \(A=\begin{bmatrix}2&3\\1&4 \end{bmatrix}\)    
\(Adj\ A=​​\begin{bmatrix}4&-3\\-1&2 \end{bmatrix}\)

Let \(|A|=\begin{vmatrix} 2&-1&-2 \\ 0 &2&-1\\3&-5&0 \end{vmatrix}\)
Expanding along R₁ we get, 
\(|A|=2\begin{vmatrix} 2 & -1\\ -5 & 0\end{vmatrix}+1\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix}-2\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}\)
= 2 ( 0 - 5 ) + 1 ( 0 + 3 )  - 2 ( 0 - 6 )
= - 10 + 3 + 12 = 5
\(\therefore\) |A| = 5.

Given \(\begin{vmatrix}2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow\)   2 - 20 = 2x² - 24
\(\Rightarrow\)  -18 = 2x² - 24
\(\Rightarrow\)  -18 + 24 = 2x²
\(\Rightarrow\)  6 = 2x²
\(\Rightarrow\) x² = 3 
\(\Rightarrow\) x = \(\pm\sqrt{3}\)

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

LHS = 8C₄+8C₃
\(=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}+\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\)
\(=70+56=126\)
\(\mathrm{RHS}=9 C_4=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}\)
\(=126\)
Hence verified

\(\frac { 7! }{ 6! } =\frac { 7\times 6! }{ 6! } =7\)

LHS 10P₃ = 10 x 9 x 8 = 720
RHS 9P₃ + 3. 9P₂ = 9 x 8 x 7 + 3 x 9 x 8
= 9 x 8 (7 + 3) = 72 (10) = 720
LHS= RHS  Hence proved.

nCx = nCy \(\Rightarrow\)x= y or x + y = n
\(\Rightarrow\)  \(\therefore\) 25 _Cn+5 = 25 C_2n-1
\(\Rightarrow\)   n + 5 = 2n-1
\(\Rightarrow\)   n =5 (or) n + 5 + 2n - 1 = 25
\(\Rightarrow\)   6 = 2n - n or 3n + 4 = 25
\(\Rightarrow\) n=  6 or 3n = 21\(\Rightarrow\) n=7
∴ n = 6 or n = 7