Class 12th

Then, KA = $k\left[\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill k{a}_{11}\hfill & \hfill k{a}_{12}\hfill & \hfill k{a}_{13}\hfill \\ \hfill k{a}_4\hfill & \hfill k{a}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill k{a}_{31}\hfill & \hfill k{a}_{32}\hfill & \hfill k{a}_{33}\hfill \end{array}\right]$

$\therefore |KA|=\left|\begin{array}{ccc}\hfill {}_{a11}\hfill & \hfill k{a}_{12}\hfill & \hfill {}_{13}\hfill \\ \hfill {}_{a_{21}}\hfill & \hfill {}_{22}\hfill & \hfill k{a}_{23}\hfill \\ \hfill {}_{31}\hfill & \hfill {}_{a_{32}}\hfill & \hfill k_{a_{33}}\hfill \end{array}\right|$

$=k{}^3\left|\begin{array}{ccc}\hfill a_{11}\hfill & \hfill a_{12}\hfill & \hfill a_{13}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{22}\hfill \\ \hfill a_{31}\hfill & \hfill a_{32}\hfill & \hfill a_{33}\hfill \end{array}\right|$

$=k^3|A|$

So, option c is correct.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Maximum voltage = 100/2 = 50V

And minimum voltage = 20/2 = 10V

Percentage modulation= max voltage -min voltage/ max voltage + min voltage $*100$

= $\frac{50-10}{50+10}*100$ = 66.67%

Peak career voltage= max voltage+ min voltage/2= 50+10/2=30V

Peak value of information voltage= 66.67/100 $*$  30= 20V

This is a Long Answer Type Questions as classified in NCERT Exemplar

height of satellite hs= 600km
As we know velocity = distance/time

 2x/4.0410^-3 = 3⁸

So x=606 km after solving 
According to Pythagoras theorem d²=x²-h²= 606²-600²= 7236

So d= 85.06km so total distance will be double from receiver to transmitter = 170.1km
And d = √2Rh
 h=7236/2*6400 = 565m

LHS = ${\left|\begin{array}{ccc}\hfill a^2+1\hfill & \hfill ab\hfill & \hfill ac\hfill \\ \hfill ab\hfill & \hfill b^2+1\hfill & \hfill bc\hfill \\ \hfill ca\hfill & \hfill cb\hfill & \hfill c^2+1\hfill \end{array}\right|}_{.}$

$=\frac{1}{abc}\left|\begin{array}{ccc}\hfill a\left(a^2+1\right)\hfill & \hfill a{b}^2\hfill & \hfill a{c}^2\hfill \\ \hfill a^{2}b\hfill & \hfill b\left(b^2+1\right)\hfill & \hfill b{c}^2\hfill \\ \hfill a^{2}c\hfill & \hfill b^{2}c\hfill & \hfill c\left(c^2+1\right)\hfill \end{array}\right|$

$=\frac{abc}{abc}\left[\begin{array}{ccc}\hfill a^2+1\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2\hfill & \hfill b^2+1\hfill & \hfill c^2.\hfill \\ \hfill a^2\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$Taking a, b&c common from R₁, R₂&R₃

$\left[\begin{array}{ccc}\hfill 1+a^2+b^2+c^2\hfill & \hfill b^2\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+1+c^2\hfill & \hfill b^2+1\hfill & \hfill c^2\hfill \\ \hfill a^2+b^2+c^2+1\hfill & \hfill b^2\hfill & \hfill c^2+1\hfill \end{array}\right]$ C_1→ C₂ + C₃.

This is a Long Answer Type Questions as classified in NCERT Exemplar

as we know that I= I_{0 $e^{-\propto x}$}

And I= 25%of I₀=

I=I₀/4

I₀/4= I_{0 $e^{-\propto x}$}

I₀ cancel from both sides

¼= $e^{-\propto x}$

Taking log on both sides log1 -log4= - $\propto x$ loge

X= log4/ $\propto$

LHS = ${\left|\begin{array}{ccc}\hfill 1+a^2-b^2\hfill & \hfill 2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab\hfill & \hfill 1-a^2+b^2\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill -2a\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|}_{.}$

$=\left|\begin{array}{ccc}\hfill 1+a^2-b^2-b(-2b)\hfill & \hfill 2ab+a(-2b)\hfill & \hfill -2b\hfill \\ \hfill 2ab-b(2a)\hfill & \hfill 1-a^2+b^2+a(2a)\hfill & \hfill 2a\hfill \\ \hfill 2b-b\left(1-a^2-b^2\right)\hfill & \hfill -2a+a\left(1-a^2-b^2\right)\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+a^2-b^2+2b^2\hfill & \hfill 2ab-2ab\hfill & \hfill -2b\hfill \\ \hfill 2ab-2ab\hfill & \hfill 1-a^2+b^2+2a^2\hfill & \hfill 2a\hfill \\ \hfill 2b-b+a^{2}b+b^3\hfill & \hfill -2a+a-a^3-a{b}^2\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+a^2+b^2\hfill & \hfill 0\hfill & \hfill -2b\hfill \\ \hfill 0\hfill & \hfill 1+a^2+b^2\hfill & \hfill 2a\hfill \\ \hfill b\left(1+a^2+b^2\right)\hfill & \hfill -a\left(1+a^2+b^2\right)\hfill & \hfill 1-a^2-b^2\hfill \end{array}\right|$

= (1 + a² + b²)² [(1 -a² + b²) - 2a (-a)]

= (1 + a² + b²)² (1 -a² + b² + 2a²)

= (1 + a² + b²)² (1 + a² + b²)

= (1 + a² + b²)³ = R.H.S.

LHS = $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x^2\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill 1+x^2+x\hfill & \hfill x+1+x^2\hfill & \hfill x^2+x+1\hfill \\ \hfill x^2\hfill & \hfill 1\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \end{array}\right|{}_{\mathrm{}}$R_1→ R₁ + R₂ + R₃

= (1 + x² + x) (1 -x)² [ (1 + x)* 1 - (-x) x].

= (1 + x² + x) (1 -x)² (1 + x + x²).

= { (1 + x² + x) (1 -x)}²

= {1 -x + x²-x³ + x-x²}²

= (1 -x³)² = R.H.S.

(i) LHS = $\left|\begin{array}{ccc}\hfill a-b-c\hfill & \hfill 2a\hfill & \hfill 2a\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill a-b-c+2b+2c\hfill & \hfill 2a+b-c-a+2c\hfill & \hfill 2a+2b+c-ab\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$R_1→ R₁ + R₂ + R₃

$=\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill a+b+c\hfill & \hfill a+b+c\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$

= (a + b + c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2b\hfill & \hfill b-c-a\hfill & \hfill 2b\hfill \\ \hfill 2c\hfill & \hfill 2c\hfill & \hfill c-a-b\hfill \end{array}\right|$ Taking (a + b + c) common from R₁

= (a + b+ c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1-1\hfill & \hfill 1-1\hfill \\ \hfill 2b\hfill & \hfill b-c-a-2b\hfill & \hfill 2b-2b\hfill \\ \hfill 2c\hfill & \hfill 2c-2c\hfill & \hfill c-a-b-2c\hfill \end{array}\right|\begin{array}{cc}\hfill {}_{}\to {}_2-{}_1\hfill & \hfill {}_{}\to {}_3-{}_1\hfill \end{array}$

= (a + b + c) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 2b\hfill & \hfill -b-c-a\hfill & \hfill 0\hfill \\ \hfill 2c\hfill & \hfill 0\hfill & \hfill -c-a-b\hfill \end{array}\right|.$

= (a + b + c) * 1. $\left|\begin{array}{cc}\hfill -(a+b+c)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -(a+b+c)\hfill \end{array}\right|$ Expand along R₁

= (a + b + c){(a + b + c)²- 0}

= (a + b +c)³ = R.H.S

(ii) LHS = ${\left|\begin{array}{ccc}\hfill x+y+2z\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{\cdot }$

$=|\begin{array}{ccc}\hfill x+y+2z+x+y\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill z+y+z+2x+y\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill z+x+z+x+2y\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}$
C_1→ C₁ + C₂ + C₃.

$={\left|\begin{array}{ccc}\hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 2(x+y+z)\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|}_{.}$

= 2 (k + y + z) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill y+z+2x\hfill & \hfill y\hfill \\ \hfill 1\hfill & \hfill x\hfill & \hfill z+x+2y\hfill \end{array}\right|$ Taking 2

(i) LHS = $\left|\begin{array}{ccc}\hfill x+4\hfill & \hfill 2x\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$

$=\left|\begin{array}{ccc}\hfill x+4+2x+2x\hfill & \hfill 2x+x+4+2x\hfill & \hfill 2x+2x+x+4\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$ R_1→R₁ + R₂ + R₃

$=\left|\begin{array}{ccc}\hfill 5x+4\hfill & \hfill 5x+4\hfill & \hfill 5x+4\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|$

Taking (5x + 4) common from R₁.

$=(5x+4)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 2x\hfill & \hfill x+4\hfill & \hfill 2x\hfill \\ \hfill 2x\hfill & \hfill 2x\hfill & \hfill x+4\hfill \end{array}\right|.$

= (5x + 4)[0 - (4 -x)(x- 4)]

= (5x + 4)(4 -x)(4 -x)

= (5x + 4)(4 -x)² = R.H.S.

= $\left|\begin{array}{ccc}\hfill y+k+y+y\hfill & \hfill y+y+k+y\hfill & \hfill y+y+y+k\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$ R_1→R₁ + R₂+ R₃

$=\left|\begin{array}{ccc}\hfill 3y+k\hfill & \hfill 3y+k\hfill & \hfill 3y+k\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$

= (3y + k) $\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y+k\hfill & \hfill y\hfill \\ \hfill y\hfill & \hfill y\hfill & \hfill y+k\hfill \end{array}\right|$

LHS $\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill yz\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill zx\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill xy\hfill \end{array}\right|$

$=\frac{1}{xyz}$ $\left|\begin{array}{ccc}\hfill x·x\hfill & \hfill x·x^2\hfill & \hfill x·yz\hfill \\ \hfill y·y\hfill & \hfill y·y^2\hfill & \hfill y·zx\hfill \\ \hfill z·z\hfill & \hfill z·z^2\hfill & \hfill z·xy\hfill \end{array}\right|$ Multiplying R1, R2& R3 by x, y&z.

$=\frac{1}{xyz}\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill xyz\hfill \\ \hfill y^2\hfill & \hfill y^3\hfill & \hfill xyz\hfill \\ \hfill z^2\hfill & \hfill z^3\hfill & \hfill xyz\hfill \end{array}\right|.$

$=\frac{xyz}{xyz}\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill y^2\hfill & \hfill y^3\hfill & \hfill 1\hfill \\ \hfill z^2\hfill & \hfill z^3\hfill & \hfill 1\hfill \end{array}\right|$ Taking xyz common from c3.

$=\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill y^2-x^2\hfill & \hfill y^3-x^3\hfill & \hfill 1-1\hfill \\ \hfill z^2-x^2\hfill & \hfill 2^3-x^3\hfill & \hfill 1-1\hfill \end{array}\right|\begin{array}{cc}\hfill {}_2\to {}_2-{}_1\hfill & \hfill {}_3\to {}_3-{}_1.\hfill \end{array}$

= $\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x^3\hfill & \hfill 1\hfill \\ \hfill (y-x)(y+x)\hfill & \hfill (y-x)\left(y^2+x^2+xy\right)\hfill & \hfill 0\hfill \\ \hfill (z-x)(z+x)\hfill & \hfill (2-x)\left(z^2+x^2+2x\right)\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along c3 we get,

= 1 $\left|\begin{array}{cc}\hfill (y-x)(y+x)\hfill & \hfill (y-x)\left(y^2+x^2+xy\right)\hfill \\ \hfill (z-x)(z+x)\hfill & \hfill (z-x)\left(z^2+x^2+zx\right)\hfill \end{array}\right|$

= (y-x)(z-x). $\left|\begin{array}{cc}\hfill y+x\hfill & \hfill y^2+x^2+xy\hfill \\ \hfill z+x\hfill & \hfill z^2+x^2+zx\hfill \end{array}\right|$ Taking (yx) & (zx) common from R₁ and R₂.

= (y-x)(z-x).  Taking (y-x) & (z-x) common from R₁ and R₂.

= (y-x)(z-x)[(y + x)(z² + x² + zx) - (z + x)(y² + x² + xy)]

= (y-x)(z-x)[yz² + yx² + xyz + xz² + x³ + x²z-zy²-zx²-xyz-xy²-x³-x²y]

= y-x)(z-x)[yz²-zy² + xz²-xy²]

= (y-x)(z-x)[yz (z-y) + x(z²-y²)]

= y-x)(z-x)(z-y)[yz + x (z + y)]             

= (- 1)(x-y)(z-x)(- 1)(y-z)[yz+ 1 xz + xy]

= (x-y)(y-z)(z-x)(xy + yz + xz). = R.H.S