(Corresponding co-factor)

(Since \(|k A|=k^n|A|,\) n is the order of matrix A

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

The supply price relation is given by
\({ x }^{ 2 }=5p-15\) 
= \(5(p-3)\) 
\(\Rightarrow { X }^{ 2 }aP\)  where X = x and \(P=p-3\) 
\(\\ \therefore \) the supply curve is a parabola whose vertex is \(\left( X=0,\ P=0 \right) \) 
i.e., The supply curve is a parabola whose vertex is (0, 3)

The given equation is
y²=4x+4y
⇒ y²-4y=4x
⇒ y²-4y=4x
⇒ y²-4y+4=4x+4 (Adding 4 on both sides)
⇒ (y-2)²=4(x+1)
⇒ y²=4 where X=x+1 ⇒ Y=y-2

The first place can be filled in only one way namely T and the remaining 7 letters can be arranged in 7! ways.
\(\therefore\) Total number of arrangements = 1 x 7! = 5040.

B \(=\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.30 \\ -0.41 & 0.67 \end{bmatrix}\)
= (0.50) (0.67) - (0.30) (0.41)
\(|I-B|\) = 0.335 - 0.123 = 0.212 >  0
Since the main diagonal elements of I - B are positive and |I-B| is positive. Hawkins Simon conditions are satisfied. Therefore given system is viable

Taking vertex of the parabola as reflector at origin, x-axis along the axis of the parabola, equation of the parabola is y² = 4ax
Given depth = 5 cm, diameter = 20 cm
\(\therefore\) (5, 10) lies on the parabola
\(\therefore\) 10² = 4a(5) ⇒ 100 = 20a ⇒ a = \(\frac{100}{20}\) = 5

\(\therefore\) Focus is (a, 0) = (5, 0) which is the mid-point of the given diameter

Now AB = \(\begin{bmatrix} 2&2&1\\1&3&1\\1&2&2 \end{bmatrix} \begin{bmatrix} {{4}\over{5}} &{{-2}\over{5}}&-{{1}\over{5}}\\-{{1}\over{5}}&{{3}\over{5}} &-{{1}\over{5}} \\ -{{1}\over{5}}&-{{2}\over{5}} &{{4}\over{5}}\end{bmatrix}\)
\(=\frac{1}{5}\left(\begin{array}{lll} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array}\right)\left(\begin{array}{ccc} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 8-2-1 & -4+6-2 & -2-2+4 \\ 4-3-1 & -2+9-2 & -1-3+4 \\ 4-2-2 & -2+6-4 & -1-2+8 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\mathrm{BA}=\frac{1}{5}\left(\begin{array}{ccc} 4 & -2 & -1 \\ -1 & 3 & -1 \\ -1 & -2 & 4 \end{array}\right)\left(\begin{array}{lll} 2 & 2 & 1 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{ccc} 8-2-1 & 8-6-2 & 4-2-2 \\ -2+3-1 & -2+9-2 & -1+3-2 \\ -2-2+4 & -2-6+8 & -1-2+8 \end{array}\right)\)
\(=\frac{1}{5}\left(\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=I\)
\(\mathrm{AB}=\mathrm{BA}=\mathrm{I}\)
\(\therefore\) A and B are inverse of each other

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁, R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)

The given equation is
\(\left( a-2 \right) { x }^{ 2 }+{ by }^{ 2 }+\left( b-2 \right) xy+4x+4y-1=0\) 
As per conditions noted above,
(i) coefficient of \(xy=0\Rightarrow b-2=0\ \therefore b=2\) 
(ii) coefficient of x² = coefficient of y²
\(\Rightarrow a-2=b\)
\(a-2=2\Rightarrow a=4\) 
\(\therefore \) Resulting equation of circle is 
\({ 2x }^{ 2 }+{ 2y }^{ 2 }+4x+4y-1=0\)

Since keys are identical, both clock wise and anti-clockwise circular permutation are same.
The number of permutation is
\(\cfrac { \left( n-1 \right) ! }{ 2 } =\cfrac { \left( 10-1 \right) ! }{ 2 } =\cfrac { 9! }{ 2 } \)

\(\Rightarrow\)   \(\alpha+2\beta={{\pi}\over{4}}\)

The circle touches the co-ordinate axes and the line x = a is shown in the diagram.
\(\therefore \ centre\ is\ \left( \frac { a }{ 2 } ,\frac { a }{ 2 } \right) and\ r=\frac { a }{ 2 } \)
\(\therefore\) There may be two such circles, one lying

above X-axis and other below X-axis.
Equation of the circle lying above the X-axis is
\({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

Equation of the circle lying below the X-axis is \({ \left( x-\frac { a }{ 2 } \right) }^{ 2 }+{ \left( y+\frac { a }{ 2 } \right) }^{ 2 }={ \left( \frac { a }{ 2 } \right) }^{ 2 }\)

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

\(y=\frac { 1 }{ 5 } { x }^{ 2 }-6x+100\)
\(y=\frac{x^2-30 x+500}{5}\)
\(\Rightarrow\)  5y = x² - 30x + 500
\(\Rightarrow\) x² - 30x = 5y - 500
\(\Rightarrow\) (x - 15)² = 5y - 500 + 225
\(\Rightarrow\) (x - 15)² = 5y - 275
\(\Rightarrow\) (x - 15)² = 5(y - 55)
\(\Rightarrow\) X² = 5Y is a parabola
where X = x - 15, Y = y - 55.
The vertex of the parabola is (15, 55) with respect to (x, y) axis. The output and the average cost at the vertex are 15 kg and Rs. 55.

\(\left(x-\frac{2}{x^2}\right)^{15}\)
\(t_{r+1}=n C_r x^{n-r} a^r\)
\(=15 C_r(x)^{15-r}\left(\frac{-2}{x^2}\right)^r\)
\(=(-1)^r 15 C_r 2^r x^{15-r}\left(\frac{1}{x^{2 r}}\right)\)
\(=(-1)^r 15 C_r 2^r x^{15-r-2 r}\)
\(=(-1)^r 15 C_r 2^r x^{15-3 r}\)
To find the term independent of x, equate the power of x to 0
15 - 3r = 0
\(3 r=15 \Rightarrow r=\frac{15}{3}=5\)
\(t_{r+1}=(-1)^5 15 C_5 2^5\)
\(=-32\left(15 C_5\right)\)