Given A is singular
\(\therefore|A|=0\)
\(\left|\begin{array}{cc} 7 & 3 \\ -2 & a \end{array}\right|=0\)
7a + 6 = 0
7a = -6
\(a=\frac{-6}{7} .\)

\(\text { (i) } \mathrm{A}=\left[\begin{array}{ll} 4 & -2 \\ 3 & -5 \end{array}\right]\)
\(We \ can \ write \ A=\frac{1}{2}\left(A+A^T\right)+\frac{1}{2}\left(A-A^T\right) (1)\)
\(A^T=\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]\)
\(A+A^{\top}=\left[\begin{array}{cc} 4 & -2 \\ 3 & -5 \end{array}\right]+\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]=\left[\begin{array}{cc} 8 & 1 \\ 1 & -10 \end{array}\right]\)
\(A-A^T=\left[\begin{array}{cc} 4 & -2 \\ 3 & -5 \end{array}\right]-\left[\begin{array}{cc} 4 & 3 \\ -2 & -5 \end{array}\right]=\left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right]\)
\(\therefore A=\frac{1}{2}\left[\begin{array}{cc} 8 & 1 \\ 1 & -10 \end{array}\right]+\frac{1}{2}\left[\begin{array}{cc} 0 & -5 \\ 5 & 0 \end{array}\right]\)

2,2,4,4,6,6
Given sequences is 2, 2, 4, 4, 6, 6,
the odd term are 2, 4, 6 .. and even terms are also 2, 4, 6
\(\therefore { \ a }_{ n= }\begin{cases} n+1 \\ 1 \end{cases}\)
if n is odd
if n is even

Given ^(n-1)P₃ :ⁿ P₄ = 1 : 10 
⇒ \(\frac { (n-1){ P }_{ 3 } }{ n{ P }_{ 4 } } =\frac { 1 }{ 10 } \)
⇒ 10.(n-1)P₃ = 1.nP₄  \(\left[ \because npr\quad =\frac { n! }{ (n-r)! } \right] \)
⇒ \(10\times \frac { (n-1)! }{ (n-1-3)! } =\frac { n! }{ (n-4)! } \)
⇒   \(\frac { 10\times (n-1)! }{ (n-4)! } =\) \(\frac { 10\times (n-1)! }{ (n-4)! }\)⇒\(\frac { n(n-1)! }{ (n-4)! } \)
⇒10 = n
∴ n = 10.

By putting sin A = sin B, we get
\(\begin{vmatrix} 1& 1 &1 \\1+sin A &1+sin B &1+sin C \\ sinA(1+sin A) &sin B(1+sin B) &sin C(1+sin C) \end{vmatrix}=0\)
That is, by putting sin A = sin B we see that, the given equation is satisfied.
Similarly by putting sin B = sin C and sin C = sin A, the given equation is satisfied.
Thus, we have A = B or B = C or C = A.
In all cases atleast two angles are equal. Thus the triangle is isosceles.

LHS = \(\begin{vmatrix} 1& a & a^2-bc \\1 &b &b^2-ca \\ 1 & c & c^2-ab \end{vmatrix}\)\(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\left| \begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{matrix} \right| \) [By proverty 7]
Multiplying and dividing R₁, R₂ and R₃ of second determinant by a, b, c respectively.
LHS \(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\frac { 1 }{ abc } \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| \)
In II determinant, Take abe from C₃
\(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\frac { abc }{ abc } \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| \)
Applying C₁ ↔️ C₃ in the second determinant,
Applying C₃ ↔️ C₂ in the second determinant
= \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \) = 0 = RHS
Hence proved.

Cost matrix A = [30 15 45], Fruit matrix

Cost of packages are obtained by computing AB. That is, by multiplying cost of each item in A (cost matrix A) with number of items in B (Fruit matrix B).
\(A B=\left[\begin{array}{lll} 30 & 15 & 45 \end{array}\right]\left[\begin{array}{lll} 6 & 5 & 6 \\ 3 & 4 & 6 \\ 3 & 4 & 6 \end{array}\right]=\left[\begin{array}{lll} 360 & 390 & 540 \end{array}\right]\)
Pack-I cost Rs. 360, Pack-II cost Rs. 390, Pack-III costs Rs. 540.

Since there is a constant increase or decrease in arithmetic progression, it is a linear function.
Let the x-axis be the number of the term and the y-axis be the value of the term.
Let (x₁, y₁) and (x₂, y₂) be (7, 30) and (10, 21) respectively, using the equation
y - y₁ = m(x -x₁) we get
y - 30 = \(\frac{21-30}{10-7}\)(x-7)
y = -3x + 51...(1)
(i) Substituting x = 1, 2 and 3 in the, equation (1) we get the first three terms of AP as 48, 45 and 42.
(ii) Substituting y = 0 in equation (1) we get 0 = - 3x + 51 ⇒ x = 17
That is seventeenth term of A.P. is zero.
(iii) Clearly the slope of the line -3 is equal to the common difference A.P.
y - y₁ = m(x - x₁) ⇒ y =y₁ + mx - mx₁
⇒ y = mx + (y₁ - mx₁)
Put x = 1, 2, 3 the values of y are m + (y₁ - mx₁), 2m + (y₁ - mx₁), 3m + (y₁ - mx₁)
They are in A.P whose common difference in m.

Let P (h, k) be any point on the required path. From the given information we have
h = a(θ - sin θ)...(1)
k = a(θ - cos θ)...(2)
Let us find the value of θ and sinθ from equation (2)
k = a(1 - cos θ)
cos θ = \(\frac{a-k}{a}\)
⇒ θ = cos^-1\((\frac{a-k}{a})\) and sin θ = \(\frac{\sqrt{2ak-k^2}}{a}\)
Substituting above values in (1) we get h = a cos^-1\((\frac{a-k}{a})-\sqrt{2ak-k^2}\)
x = a cos^-1\((\frac{a-y}{a})-\sqrt{2ay-y^2}\)

⁽ⁿ⁺²⁾C₇ : ^(n-1)P₄ = 13 : 24
\(\frac { { (n+2) }_{ C_{ 7 } } }{ { (n+2) }_{ P_{ 4 } } } =\frac { 13 }{ 24 } \)
\(\frac { (n+2)! }{ (n-5)!7! } \times \frac { (n-5)! }{ (n-1)! } =\frac { 13 }{ 24 } \)
\(\frac { (n+2)(n+1)n(n-1)! }{ (n-1)!.7! } =\frac { 13 }{ 24 } \)
\((n+2)(n+1)(n)=\frac{13}{24} \times 7 !=\frac{13}{24} \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
\((n+2)(n+1)(n)=13 \times 14 \times 15\)
n + 2 = 15 \(\Longrightarrow\) n = 13.

(i) Let the variable x represents the quantity of gas, and y represents the number of days.
By the given data,

Using two - point form, the linear relationship between quantity of gas in the cylinder to the number of days is
\(\frac { y-0 }{ 24-0 } =\frac { x-0 }{ 14.2-0 } \ \left[ \because \ \frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\(\frac { y }{ 24 } =\frac { x }{ 14.2 } \)
\(\Rightarrow\)   14.2 y = 24x
\(\Rightarrow\)  24x - 14.2y = 0
\(\Rightarrow\)  12x - 7.1y = 0  ....(1)
Which is the required linear relation...(2)
Find the equation relating the quantity of gas in the cylinder to the days.
To find the Time taken to cross the pole, put y = 0
\(\Rightarrow \ \therefore \ 12.5x\ =\ 150\ \ \Rightarrow \ x=\frac { 150 }{ 12.5 } =12\ sec\)
(ii) Draw the graph for first 96 days

y = f(x) is a periodicfunctionwith period 24. 
Therefore, f(x) = f(x + 24)

(i) Let the x - axis represent the weight and the y-axis represent the length

(ii) Consider the point \(\begin{pmatrix} { x }_{ 1 } & { y }_{ 1 } \\ 2 & 3 \end{pmatrix}\begin{pmatrix} { x }_{ 2 } & { y }_{ 2 } \\ 4 & 4 \end{pmatrix}\)
Equation of the straight line using two point form is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\Rightarrow \quad \frac { y-3 }{ 4-3 } =\frac { x-2 }{ 4-2 } \)
\(\Rightarrow \quad \frac { y-3 }{ 1 } =\frac { x-2 }{ 2 } \)
\(\Rightarrow\) 2y - 6 = x-2
 \(\Rightarrow\) x - 2y + 4 = 0
(iii) To find the actual length of the spring,
Put x = 0
\(\Rightarrow\) 0 - 2y + 4 = 0
\(\Rightarrow\) -2y = -4
\(\Rightarrow\) y = 2 cm
(iv) Put y = 9 in (1) we get
\(\Rightarrow\) x - 18 + 4 = 0
\(\Rightarrow\) x - 14 = 0
\(\Rightarrow\) x = 14 kg
\(\therefore\) 14 Kg must be added
(v) Put x = 6 in (1) we get,
\(\Rightarrow\)  6 - 2y + 4 = 0
\(\Rightarrow\) 10 - 2y = 0
\(\Rightarrow\) 10 = 2y 
\(\Rightarrow\) y = 5
\(\therefore\) Strength of the spring = 5 cm

Since 0 cannot be in the thousand's place
Let us find the sum of all these 500 numbers.
By filling 0 is the unit place, the remaining 3 places can be filled with remaining 4 digits is \(4\times 4\times 4=64\)  in unit place.
∴ Sum of all the unit digits
= \((64\times 0)+(64\times 2)+(64\times 5)+(64\times 7)+(64\times 8)\)
= 64(0 + 2 + 5 + 7 + 8) = 64 (22) = 1408
Similarly sum of the tens digits =\(14408\times 10\) = 144080
Sum of all the hundred's digits =  \(14408\times 100\) = 1440800
By filling 2 in the thousand's place, remaining 3 places can be filled with remaining 3 digits in \(3\times 3\times 3\) 27 ways [since 0 cannot be in thousands place]
∴ Sum of the digits in the thousands place
= \(27(2+5+7+8)=27(22)\times 1000\) = 594000
Hence sum of all the 4-digit numbers formed by using the digits 0, 2,5, 7, 8 is 1408 + 14080 + 140800 + 594000
= 1408 (1 + 10 + 100) + 594000
= \(1408\times 11+594000\)
= 750288

\(A_{m \times n}\left[\begin{array}{ccc} 1, & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]_{2 \times 3}=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \end{array}\right]_{2 \times 3}\)
\(n=2 \\ m=2\ [Since \ 2 \times 2.2 \times 3 \Rightarrow 2 \times 3 ]\\ \therefore A \ is \ of\ order \ 2 \times 2\)
Let us take
\(A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\)
\(\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \end{array}\right]\)
\(\left[\begin{array}{ccc} a+4 b & 2 a+5 b & 3 a+6 b \\ c+4 d & 2 c+5 d & 3 c+6 d \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \end{array}\right]\)
\(a+4 b=-7 ..........(1)\\ 2 a+5 b =-8 .........(2)\\ c+4 d=2 ............(3)\\ 2 c+5 d =4 ...........(4)\)
\(\text { (1) } \times 2 \Rightarrow 2 a+8 b=-14\)
\(\text { (2) } x-1 \Rightarrow-2 a-5 b=8\)
3b = -6
b = -2
Put in (1)
\(a+4(-2)=-7\)
\(a-8=-7\)
\(a=-7+8\)
a = 1
\(2 \times(3) \Rightarrow 2 c+8 d=4\)
\(-2 c-5 d=-4\)
\(3 \mathrm{~d} =0 \)
d = 0
Put in (3)
C + 0 2
C = 2
\(\therefore A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{cc} 1 & -2 \\ 2 & 0 \end{array}\right]\)

7 + 77 + 777 + 7777 + ...
S_n = 7 + 77 + 777 + 7777 + ... to n terms
= \(\frac{7}{9}[9+99+999+9999+\)...to n terms]
= \(\frac{7}{9}[(10-1)+(10^{2}-1)+(10^{3}-1)+(10^{4}-1)+..\).to n terms]
= \(\frac{7}{9}[(10+10^{2}+10^{3}+... to n terms)-(1+1+1+... n terms)]\)
= \(\frac{7}{9}[\frac{10(10^{n}-1)}{10-1}-n]=\frac{7}{9}[\frac{10}{9}(10^{n}-1)-n]\)
= \(\frac{7}{81}[10^{n+1}-9n-10]\)

(a) Total number of arrangement when boys and girls alternate = (5!)² + (5!)²
(b) No two girls sit together = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together = 10! - 5! 6!
Hence, the required matching is
(a)\(\leftrightarrow\)(iii), (b)\(\leftrightarrow\)(i), (c) \(\leftrightarrow\)(iv), (d)\(\leftrightarrow\)(ii)

When n = 18 (even)
a_n = n (n + 2) = 18 (18 + 2) = 18 (20) = 360
When n = 25 (odd)
\(a_{n}=\frac{4n}{n^{2}+1}=\frac{4(25)}{(25)^{2}+1}=\frac{100}{625+1}=\frac{100}{626}=\frac{50}{313}\)

Here n = \(2\over 3\)
\({n(n-1)\over 2!}={{2\over3}({2\over3})-1\over 2!}={{2\over3}({-1\over3})\over 2}={-1\over9}\)
\({n(n-1)(n-1)\over 2!}={{2\over 3}({2\over 3}-1)({2\over 3}-2)\over 3!}={{2\over 3}({-1\over 3})({-4\over 3})\over 6}={4\over 18}\)
Thus(1+x)^\(2\over3\) = \(1+{2\over3}x-{1\over 9}x^2+{4\over 81}x^3+.....\)

Given 3 + 10 + 17 +...
a₁ = 3, d₁ = 10 - 3 = 7
∴ T_n = a₁(n-1)d₁ = 3 + (n-1)7 = 7n-4  ...(1)
Also, given 63 + 65 + 67+..
a₂ = 63, d₂ = 65 - 63 = 2
∴ T_n = a₂+(n-1)d₂ = 63 + (n-1) 2 = 2n + 61   ..(2)
Let n^th term of given series be equal
⇒   7n - 4 =2n + 61
⇒ 5n = 65     [From (1)and (2)]
⇒ n = 13