We shall prove the result by using principle of mathematical induction.
Let \(P\left( n \right) { :A }^{ n }=\left[ \begin{matrix} { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \\ { 3 }^{ n-1 } & { 3 }^{ n-1 } & { 3 }^{ n-1 } \end{matrix} \right] \)
Now, \(P\left( 1 \right) { :A }^{ 1 }=\left[ \begin{matrix} { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \\ { 3 }^{ 0 } & { 3 }^{ 0 } & { 3 }^{ 0 } \end{matrix} \right] =\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \)
The result is true for n = 1.
Let the result be true for n = k.
So, \({ A }^{ k }=\left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
Now, we prove that P(k + 1) is true.
Now, A^k+1 = A. A^k
\(=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \\ { 3 }^{ k-1 } & { 3 }^{ k-1 } & { 3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \\ { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } & { 3.3 }^{ k-1 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \\ { 3 }^{ k } & { 3 }^{ k } & { 3 }^{ k } \end{matrix} \right] \)
= A^k+1
Hence, it is true n = k + 1.
Hence, by principle of mathematical induction P(n) is true for all \(n\epsilon N\)

We know that, \(A \cup B\) denotes the occurrence of at least one of A and B and \(A \cap B\) denotes the occurrence of both A and B, simultaneously.
Then, \(P(A \cup B)=0.6 \text { and } P(A \cap B)=0.3\) 
\(\because P(A \cup B)=P(A)+P(B)-P(A \cap B)\)
\(\therefore \quad 0.6=P(A)+P(B)-0.3 \Rightarrow P(A)+P(B)=0.9\) 
\(\Rightarrow\{[1-P(\bar{A})]+[1-P(\bar{B})]\}=0.9\) 
\([\because P(A)=1-P(\bar{A}) \text { and } P(B)=1-P(\bar{B})]\) 
\(\Rightarrow P(\bar{A})+P(\bar{B})=2-0.9=1.1\) 

Let the equation of line passing through the point (3, 0, 1) is
\(\frac{x-3}{a}=\frac{y-0}{b}=\frac{z - 1}{c}\) where a, b, c are DR's of the line.
Now, as this line is parallel to the plane
\(x+2 y=0 \text { and } 3 y-z=0\) 
Therefore, we have
a ·1 + b - 2 + c .0 = 0, i.e. a + 2b = 0...(i)
and a·0 + b·3 + c·(-1) = 0, i.e. 3b - c = 0...(ii)
From Eq. (ii), we get \(b=\frac{c}{3}\) .
\(a=-2 b=-2\left(\frac{c}{3}\right)=\frac{-2 c}{3}\) [usingEq. (iii)]..(iv)
Hence, the required equation offline is given by
\(\frac{x-3}{\left(\frac{-2 c}{3}\right)}=\frac{y}{\frac{c}{3}}=\frac{z-1}{c}\) [using Eqs. (ill) and (iv)]
\(\Rightarrow \frac{3(x-3)}{-2}=\frac{3 y}{1}=\frac{z-1}{1}\) 
\(\Rightarrow \frac{(x-3)}{-2}=\frac{y}{1}=\frac{z-1}{3}\) [divide each term by 3]

Given equation of plane is x + y + z = 1
On dividing both sides by \(\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{3}\) 
we get
\(\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}} y+\frac{1}{\sqrt{3}} z=\frac{1}{\sqrt{3}}\) 
which is the form of \(l x+m y+n z=d\) 
\(\left[\because l^{2}+m^{2}+n^{2}=\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\right]\) 
\(\therefore\) Direction cosines \((l, m, n)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\)  and distance from the origin \(d=\frac{1}{\sqrt{3}} \text { units. }\)

\(\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k})\)

\((5,-2,0) ;(5 \hat{i},-2 \hat{j}, 0 \hat{k})\)

(i) \(\int e^{2 y} d y=\int 1 d x\) 
(ii) Put y = 3, to get the required value of 
\(=\frac{e^{6}+9}{2}\)

Let \(I=\int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}=\int_{0}^{1} \frac{e^{x}}{e^{2 x}+1} d x\) 
Now, put \(e^{x}=t \Rightarrow e^{x} d x=d t\) 
Upper limit When x = 1, then t = e¹ = e
Lower limit When x = 0, then t = e^o = 1
\(\therefore \ I=\int_{1}^{e} \frac{d t}{t^{2}+1}=\left[\tan ^{-1} t\right]_{1}^{e}\) 
\(=\tan ^{-1} e-\tan ^{-1} 1=\tan ^{-1} e-\frac{\pi}{4}\) 

On putting sin x = t, given integral reduces to
\(\int_{0}^{1} e^{t} d t .[\text { Ans. } e-1]\)

Let \(I=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{\circ}}-e^{\log x^{3}}}{e^{\log x^{4}}-e^{\log x^{3}}} d x\) 
\(=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x \quad\left[\because e^{\log f(x)}=f(x)\right][1]\)
\(=\int \frac{x^{5}(x-1)}{x^{3}(x-1)} d x=\int x^{2} d x=\frac{x^{3}}{3}+C \) 

Given is y \(y=5 x-2 x^{3} \text { and } \frac{d x}{d t}=2 \text { units } / \mathrm{s}\) 
Now, slope of the curve \(\frac{d y}{d x}=5-6 x^{2}=M(\text { say })\) 
Rate of change of the slope
\( \frac{d M}{d t} =-6 \frac{d}{d t}\left(x^{2}\right) \)
\( \Rightarrow \frac{d M}{d t} =-12 x \frac{d x}{d t} \) 
When x = 3 , then
\(\frac{d M}{d t}=-12 \times 3 \times 2=-72 \text { unit/s }\) 
Thus, the slope of decreasing at a rate of 72 units/s.

\(\frac{d V}{d t}=k \text { (constant) }\) 
\( \Rightarrow 3 a^{2} \frac{d a}{d t} =k \Rightarrow \frac{d a}{d t}=\frac{k}{3 a^{2}} \)
\(\text { Now, } \ \frac{d S}{d t} =\frac{d}{d t}\left(6 a^{2}\right)=12 a \frac{d a}{d t} \)
\(=12 a \cdot \frac{k}{3 a^{2}}=4 \frac{k}{a} \Rightarrow \frac{d S}{d t} \propto \frac{1}{a} \)

Let r be the radius, A be the area and P be the perimeter.
Then, we have \(\frac{d A}{d t}=\text { constant }=k(\text { say })\)

\(\begin{array}{l} f(x)=-\cos x, \text { if } \frac{\pi}{2}

Given IAI = 12.
Determinant formed by cofactors is (adj A)'
Also I(adj A)'I = I(adj A)I = IAI2 [ IAI is of order 3]
= (12)² = 144.

A=0

We have,\(\begin{equation} \Delta=\left|\begin{array}{rrr} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a \end{array}\right|=86 \end{equation}\) 
\(\Rightarrow\) 1(2a² +4)- 2(-4a- 20)+0=8
[expanding along first column]
\(\Rightarrow 2 a^{2}+4+8 a+40=86\)
\(\Rightarrow\ 2 a^{2}+8 a+44-86=0\)
\(\Rightarrow a^{2}+4 a-21=0\)
\(\Rightarrow a^{2}+7 a-3 a-21=0\)
\(\Rightarrow (a+7)(a-3)=0\)
\(\Rightarrow a=-7\ and\ 3\) 
\(\therefore\) Required sum = -7 + 3 = - 4

Now,
\(A B =\left[\begin{array}{cc} 2 & 0 \\ 1 & 4 \end{array}\right]\left[\begin{array}{rr} -1 & 2 \\ 3 & 0 \end{array}\right] \)
\(=\left[\begin{array}{cc} -2+0 & 4+0 \\ -1+12 & 2+0 \end{array}\right]=\left[\begin{array}{cc} -2 & 4 \\ 11 & 2 \end{array}\right] \) 
[multiplying rows by columns]
\(\therefore \ (A B)^{\prime}=\left[\begin{array}{rr} -2 & 4 \\ 11 & 2 \end{array}\right]=\left[\begin{array}{rr} -2 & 11 \\ 4 & 2 \end{array}\right]\) 
[interchanging the elements of rows and columns]

Given matrix equation is
\( \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]=O \)
\(\Rightarrow \left[\begin{array}{ll} 2 x & 3 \end{array}\right]\left[\left[\begin{array}{rr} 1 & 2 \\ -3 & 0 \end{array}\right]\left[\begin{array}{l} x \\ 8 \end{array}\right]\right)=O \) 
[by associative law of multiplication]
\(\Rightarrow \left[\begin{array}{ll}2 x & 3\end{array}\right]\left[\begin{array}{c}x+16 \\ -3 x\end{array}\right]=0\)
\(\Rightarrow [2 x(x+16)-9 x]=[O]\)
\(\Rightarrow\) 2 x^{2} + 32 x - 9x = 0
\(\Rightarrow\)\(2 x^{2}+23 x=0\)
\(\Rightarrow x(2 x+23)=0\)
\(\Rightarrow\) x = 0 and x = -23 / 2

Given, AB = BA
Now, (A + B)² = (A + B)·(A + B)
= A·(A + B) + B·(A + B)
= A² + AB + BA + B²
= A² + AB + AB + B² [ஃ BA = AB, given]
= A ² + 2AB + B²

(i) 3 x 3
(ii) 9
(iii)a₂₃ = x² - y, a31 = 0, a₁₂ = 1]

\(\text {Here, } \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}, x y<1\)
\(\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4}\)
\(\frac{x+y}{1-x y}=1\)
\( x+y=1-x y\)
\(x+y+x y=1\)
Therefore, the value of \( x+y+x y \text { is } 1 \text { . } \)