Maths Solve the Following Question.(5 Marks)

$
\begin{gathered}
=\frac{1}{a}\left|\begin{array}{lll}
a & a^2 & a b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right| \\
\mathrm{R}_2 \rightarrow \mathrm{bR}_2
\end{gathered}
$
$=\frac{1}{a} \times \frac{1}{b}\left|\begin{array}{lll}a & a^2 & a b c \\ b & b^2 & a b c \\ 1 & c & a b\end{array}\right|$
$
\mathrm{R}_3 \rightarrow \mathrm{cR}_3
$
$=\frac{1}{a} \times \frac{1}{b} \times \frac{1}{c}\left|\begin{array}{lll}a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c\end{array}\right|$
$=\frac{1}{a b c} \times a b c\left|\begin{array}{lll}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{array}\right|$
(taking abc common from $\mathrm{C}_3$ )
$
\begin{aligned}
& =\left|\begin{array}{lll}
a & a^2 & 1 \\
b & b^2 & 1 \\
c & c^2 & 1
\end{array}\right| \\
& \mathrm{C}_1 \leftrightarrow \mathrm{C}_3 \\
& =(-1)\left|\begin{array}{lll}
1 & a^2 & a \\
1 & b^2 & b \\
1 & c^2 & c
\end{array}\right| \\
& \mathrm{C}_2 \leftrightarrow \mathrm{C}_3 \\
&
\end{aligned}
$
$
\begin{gathered}
=(-1)\left|\begin{array}{lll}
1 & a^2 & a \\
1 & b^2 & b \\
1 & c^2 & c
\end{array}\right| \\
\mathrm{C}_2 \leftrightarrow \mathrm{C}_3
\end{gathered}
$
$=(-1)(-1)\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$
$
=\left|\begin{array}{lll}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2
\end{array}\right|=\text { R.H.S. }
$

Let $P(n)=A^{\mathbb{n}}=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$, for all $n \in N$.

Step 1: Put $n=1$

$\therefore \quad$ R.H.S. $=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & \cos \theta\end{array}\right]=\mathbf{A}=$ L.H.S.

$\therefore \quad P(n)$ is true for $n=1$.

Step 2: Let us consider that $\mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=\mathrm{k}$.

$\therefore \quad A^k=\left[\begin{array}{cc}\cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta\end{array}\right]$

...(i)

Step 3: We have to prove that $P(n)$ is true for

$\mathrm{n}=\mathrm{k}+\mathrm{l}$

i.e., to prove that

$\begin{aligned} & A^{k+1}=\left[\begin{array}{cc}\cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta\end{array}\right] \\ & \text { R.H.S. }=\left[\begin{array}{cc}\cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta\end{array}\right] \\ & \text { L.H.S. }=A^{k+1}=A^k \cdot A\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}\cos k \theta & \sin k \theta \\ -\sin k \theta & \cos k \theta\end{array}\right]\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right] \quad \ldots[\text { From (i)] } \\ & =\left[\begin{array}{cc}\cos k \theta \cos \theta-\sin k \theta \sin \theta & \cos k \theta \sin \theta+\sin k \theta \cos \theta \\ -\sin k \theta \cos \theta-\cos k \theta \sin \theta & -\sin k \theta \sin \theta+\cos k \theta \cos \theta\end{array}\right] \\ & =\left[\begin{array}{cc}\cos k \theta \cos \theta-\sin k \theta \sin \theta & \sin k \theta \cos \theta+\cos k \theta \sin \theta \\ -(\sin k \theta \cos \theta+\cos k \theta \sin \theta) & \cos k \theta \cos \theta-\sin k \theta \sin \theta\end{array}\right]\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}\cos (k \theta+\theta) & \sin (k \theta+\theta) \\ -\sin (k \theta+\theta) & \cos (k \theta+\theta)\end{array}\right] \\ & =\left[\begin{array}{cc}\cos (k+1) \theta & \sin (k+1) \theta \\ -\sin (k+1) \theta & \cos (k+1) \theta\end{array}\right]\end{aligned}$

$=$ R.H.S.

$\therefore \quad P(n)$ is true for $n=k+1$.

Step 4: From all steps above, by the principle of

Mathematical induction, $P(n)$ is true for all

$\mathrm{n} \in \mathrm{N}$.

$\therefore \quad A^n=\left[\begin{array}{cc}\cos n \theta & \sin n \theta \\ -\sin n \theta & \cos n \theta\end{array}\right]$ for all $n \in N$.

Let $P(n) \equiv A^n=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$, for all $n \in N$.

Step 1: Put n $=1$

$\begin{aligned} \text { R.H.S. } & =\left[\begin{array}{cc}1+2(1) & -4(1) \\ 1 & 1-2(1)\end{array}\right] \\ & =\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right] \\ & =\mathrm{A}=\text { L.H.S. }\end{aligned}$

$\therefore \quad \mathrm{P}(\mathrm{n})$ is true for $\mathrm{n}=1$.

Step 2: Let us consider that $P(n)$ is true for $n=k$.

$\therefore \quad A^k=\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]$

...(i)

Step 3: We have to prove that $P(n)$ is true for

$\mathrm{n}=\mathrm{k}+1$

i.e., to prove that

$\begin{aligned} \mathrm{A}^{\mathrm{k}+1} & =\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right] \\ \text { R.H.S. } & =\left[\begin{array}{cc}1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1)\end{array}\right] \\ & =\left[\begin{array}{cc}1+2 k+2 & -4 k-4 \\ k+1 & 1-2 k-2\end{array}\right] \\ & =\left[\begin{array}{cc}3+2 k & -4 k-4 \\ k+1 & -2 k-1\end{array}\right]\end{aligned}$

$\begin{aligned} \text { L.H.S. } & =A^{k+1} \\ & =A^k \cdot A\end{aligned}$

$\begin{aligned} & =\left[\begin{array}{cc}1+2 k & -4 k \\ k & 1-2 k\end{array}\right]\left[\begin{array}{cc}3 & -4 \\ 1 & -1\end{array}\right] \ldots[\text { From (i)] } \\ & =\left[\begin{array}{cc}3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k\end{array}\right] \\ & =\left[\begin{array}{cc}3+2 k & -4 k-4 \\ k+1 & -2 k-1\end{array}\right]=\text { R.H.S. }\end{aligned}$

$\therefore \quad P(n)$ is true for $n=k+1$.

Step 4: From all steps above, by the principle of

Mathematical induction $P(n)$ is true for all

$\mathbf{n} \in \mathbf{N}$.

$\therefore \quad A^n=\left[\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right]$ for all $n \in N$.

That is $\mathrm{P}(\mathrm{n})$ is true $\rightarrow \mathrm{P}(2)$ is true $\rightarrow \mathrm{P}(3)$ is true and so on $\rightarrow P(n)$ is true, $n \in N$.
$\therefore=\mathrm{A}^n\left[\begin{array}{cc}a^n & 0 \\ 0 & b^n\end{array}\right] \quad$ for all $\mathrm{n} \in \mathrm{N}$.

2.Both book shops got 10% profit in the month of August 2017.For Suresh:

Profit for Physics books $=\frac{6650 \times 10}{100}=₹ 665$

Profit for Chemistry books $=\frac{7055 \times 10}{100}=₹ 705.50$

Profit for Mathematics books $=\frac{8905 \times 10}{100}=₹ 890.50$

For Ganesh:

Profit for Physics books $=\frac{7000 \times 10}{100}=₹ 700$

Profit for Chemistry books $=\frac{7500 \times 10}{100}=₹ 750$

Profit for Mathematics books $=\frac{10200 \times 10}{100}=₹ 1020$

[Note: Answers given in the textbook for Suresh’s profit in Chemistry and Mathematics books are ? 675 and ?850 respectively. However, as per our calculation profit amounts are ₹ 705.50 and ₹ 890.50 respectively.]

$=\left[\begin{array}{cc}2-1 & -3+2 \\ 5+2 & -4+2 \\ -6+0 & 1+3\end{array}\right]$

$\therefore \quad A+B=\left[\begin{array}{cc}1 & -1 \\ 7 & -2 \\ -6 & 4\end{array}\right]$

....(i)

$B+A=\left[\begin{array}{cc}-1 & 2 \\ 2 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]$

$=\left[\begin{array}{cc}-1+2 & 2-3 \\ 2+5 & 2-4 \\ 0-6 & 3+1\end{array}\right]$

$\therefore \quad B+A=\left[\begin{array}{cc}1 & -1 \\ 7 & -2 \\ -6 & 4\end{array}\right]$

From (i) and (ii), we get A + B = B + A

$(A+B)+C=\left\{\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]+\left[\begin{array}{cc}-1 & 2 \\ 2 & 2 \\ 0 & 3\end{array}\right]\right\}+\left[\begin{array}{cc}4 & 3 \\ -1 & 4 \\ -2 & 1\end{array}\right]$

$=\left[\begin{array}{cc}2-1 & -3+2 \\ 5+2 & -4+2 \\ -6+0 & 1+3\end{array}\right]+\left[\begin{array}{cc}4 & 3 \\ -1 & 4 \\ -2 & 1\end{array}\right]$

$=\left[\begin{array}{cc}1 & -1 \\ 7 & -2 \\ -6 & 4\end{array}\right]+\left[\begin{array}{cc}4 & 3 \\ -1 & 4 \\ -2 & 1\end{array}\right]$

$=\left[\begin{array}{cc}1+4 & -1+3 \\ 7-1 & -2+4 \\ -6-2 & 4+1\end{array}\right]$

$\therefore \quad(\mathrm{A}+\mathrm{B})+\mathrm{C}=\left[\begin{array}{cc}5 & 2 \\ 6 & 2 \\ -8 & 5\end{array}\right]$

$\ldots$...(i)

$\begin{aligned} A+(B+C) & =\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]+\left\{\left[\begin{array}{cc}-1 & 2 \\ 2 & 2 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}4 & 3 \\ -1 & 4 \\ -2 & 1\end{array}\right]\right\} \\ & =\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]+\left[\begin{array}{cc}-1+4 & 2+3 \\ 2-1 & 2+4 \\ 0-2 & 3+1\end{array}\right] \\ & =\left[\begin{array}{cc}2 & -3 \\ 5 & -4 \\ -6 & 1\end{array}\right]+\left[\begin{array}{cc}3 & 5 \\ 1 & 6 \\ -2 & 4\end{array}\right]\end{aligned}$

$=\left[\begin{array}{cc}2+3 & -3+5 \\ 5+1 & -4+6 \\ -6-2 & 1+4\end{array}\right]$

$=\left[\begin{array}{cc}5 & 2 \\ 6 & 2 \\ -8 & 5\end{array}\right]$

..(ii)

$(A+B)+C=A+(B+C)$

$D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -1 & 1 \\ 2 & 1 & -1\end{array}\right|$

= 1(1 – 1) – 1(-1 – 2) + 1(1 + 2) = 1(0) – 1(-3) + 1(3) = 0 + 3 + 3 = 6 ≠ 0

$D_x=\left|\begin{array}{ccc}15 & 1 & 1 \\ 5 & -1 & 1 \\ 4 & 1 & -1\end{array}\right|$

$\begin{aligned} & =15(1-1)-1(-5-4)+1(5+4) \\ & =15(0)-1(-9)+1(9) \\ & =0+9+9 \\ & =18\end{aligned}$

$D_y=\left|\begin{array}{ccc}1 & 15 & 1 \\ 1 & 5 & 1 \\ 2 & 4 & -1\end{array}\right|$

= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10) = 1(-9) – 15(-3) + 1(-6) = -9 + 45 – 6 = 30

$D_z=\left|\begin{array}{ccc}1 & 1 & 15 \\ 1 & -1 & 5 \\ 2 & 1 & 4\end{array}\right|$

= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2) = 1(-9) – 1(-6) + 15(3) = -9 + 6 + 45 = 42

By Cramer’s Rule,

$x=\frac{\mathrm{D}_x}{\mathrm{D}}=\frac{18}{6}=3, y=\frac{\mathrm{D}_y}{\mathrm{D}}=\frac{30}{6}=5$

$z=\frac{D_z}{D}=\frac{42}{6}=7$

∴ The three numbers are 3, 5 and 7.