\(\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x} \)
\(\text { Put } x=\tan \theta \Rightarrow \theta=\tan ^{-1} x \)
\(\therefore \tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right) \)
\(=\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \)
\(=\tan ^{-1}\left(\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \)
\(=\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x \)

Let \(A=\left[\begin{array}{rrr} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right]\) 
 \(|A|=\left|\begin{array}{rrr} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{array}\right|\) 
\(\begin{aligned} =(-1)^{1+1}(3)\left|\begin{array}{rr} 1 & -2 \\ 3 & 1 \end{array}\right|+(-1)^{1+2}(-4) &\left|\begin{array}{rr} 1 & -2 \\ 2 & 1 \end{array}\right| +(-1)^{1+3}(5)\left|\begin{array}{ll} 1 & 1 \\ 2 & 3 \end{array}\right| \end{aligned}\) 
[expanding along R₁]
\(=(-1)^{2} 3[1 \times 1-3 \times(-2)]+(-1)^{3}(-4)[1 \times 1-2 \times(-2)]\)+(-1)⁴5(3- 2)
= 3(1+6)+4(1+4)+5(1)
= 21+ 20 + 5 = 46

\( \sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\} \)
\({\left[\because \frac{2 \pi}{3} \notin\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \text { so we write } \frac{2 \pi}{3} \text { as }\left(\pi-\frac{\pi}{3}\right)\right]} \)
\(=\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \quad[\because \sin (\pi-\theta)=\sin \theta] \)
\(=\frac{\pi}{3} \quad\left[\quad \because \frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\right] \)

Set A is the set of all books in the library of a college.
R = {x, y): x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.
Let (x, y) ∈ R ⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈R and (y, z) ∈ R.
⇒ x and y and have the same number of pages and y and z have the same number of pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.

We have: \(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\) 
\(x\begin{bmatrix} 2x & 2 \\ 3 & x \end{bmatrix}+2\begin{bmatrix} 8 & 5x \\ 4 & 4x \end{bmatrix}=2\begin{bmatrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{bmatrix}.\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{bmatrix}+\begin{bmatrix} 16 & 10x \\ 8 & 8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
\(\Rightarrow \ \begin{bmatrix} { 2x }^{ 2 }+16 & 12x \\ 3x+8 & { x }^{ 2 }+8x \end{bmatrix}=\begin{bmatrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 12x \end{bmatrix}\)
Comparing, \(12x=48,\quad 3x+8=20,\quad and\quad { x }^{ 2 }+8x=12x.\) 
All these give \(x=4\)

We have: \(\quad A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
\(\therefore { A }^{ 2 }=AA\\ =\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
\(=\begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix}=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}.\)
Now    \({ A }^{ 2 }-5A+7I=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}+7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
\(=\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}-\begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}+\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
\(=\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}\)
\( =\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}=0.\)

(i)\(\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow 2 \times 1-5 \times 4=2 x \times x-6 \times 4 \)
\(\Rightarrow 2-20=2 x^{2}-24 \)
\(\Rightarrow 2 x^{2}=6 \)
\(\Rightarrow x^{2}=3 \)
\(\Rightarrow x=\pm \sqrt{3} \)
(ii)We have:
\(\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\2x & 5 \end{vmatrix}\)
\(\Rightarrow 2 \times 5-3 \times 4=x \times 5-3 \times 2 x \)
\(\Rightarrow 10-12=5 x-6 x \)
\(\Rightarrow-2=-x \)
\(\Rightarrow x=2\)

3*4 = 2(3) + 4 - 3 = 6 + 4 - 3 = 7

\(|A| =\left|\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \)
\(=2(0)+3(-2)+5(1)=-1 \neq 0 \)
\(\text {Hence, } A^{-1} \text { exists. }\)
\(\text {Cofactors of elements of }|A| \text { are }\)
\(A_{11} =0, A_{12}=2, A_{13}=1 \)
\(A_{21} =-1, A_{22}=-9, A_{23}=-5 \)
\(A_{31} =2, A_{32}=23, A_{33}=13 \)
\(\operatorname{adj} A =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{T} \)
\(=\left[\begin{array}{lll} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(A^{-1} =\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{1}\left[\begin{array}{ccc} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]\)
\(\text {Consider equations, }\)
\(2 x-3 y+5 z=11 \)
\(3 x+2 y-4 z=-5 \)
\(x+y-2 z=-3\)
Corresponding matrix equation is
\(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 11 \\ -5 \\ -3 \end{array}\right]\)
\(\text { i.e., } \ A X=B \text { , Its solution is } X=A^{-1} B\)
\(X=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]\left[\begin{array}{r} 11 \\ -5 \\ -3 \end{array}\right] \quad \text { [from{i} ] }\)
\(\Rightarrow \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 0-5+6 \\ -22-45+69 \\ -11-25+39 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right] \)
\(\Rightarrow x=1, y=2, z=3 \text { is solution. } \)

We have a function \(f: R^{+} \rightarrow R^{+}\), defined as \(f(x)=e^{x}\)
For one-one Let \(x_{1}, x_{2} \in R^{+}\) such that
\(f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow e^{x_{1}}=e^{x_{2}}\)
On taking log with base e on both sides, we get
\(\log e^{x_{1}}=\log e^{x_{2}}\)
\(\Rightarrow\) \(x_{1}=x_{2} \quad[\because \log e=1]\)
So, f is one-one. For onto Let \(y \in R^{+}\) be any arbitrary element. Now, let y = f(x) \(\Rightarrow y=e^{x} \Rightarrow \log y=\log e^{x}\)
\(\Rightarrow x=\log y \in R^{+}, \forall y \in R^{+}-(0,1]\)
Thus, for \(y \in(0,1]\) , there does not exist any \(x \in R^{+}\) such that \(f(x)=y\) . So, f is not onto.
Hence, f is not invertible.

Note that the characteristic of sets {1, 4, 7}, {2, 5, 8} and {3, 6, 9} is that difference between any two elements of these sets is a multiple of 3.
Therefore, (x, y) \(\in\) R₁ \(\Rightarrow\) x – y is a multiple of 3 \(\Rightarrow\) {x, y} \(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) (x, y) \(\in\) R₂.
Hence, R₁ \(\subset\) R₂.
Similarly, {x, y} \(\in\) R₂ Þ {x, y}\(\subset\) {1, 4, 7} or {x, y} \(\subset\) {2, 5, 8} or {x, y} \(\subset\) {3, 6, 9} \(\Rightarrow\) x – y is divisible by 3 \(\Rightarrow\) {x, y} \(\in\) R₁.
This shows that R₂ \(\subset\) R₁.
Hence, R₁ = R₂

Getting,\(\sin ^{ -1 }{ \left( \frac { 5 }{ 13 } \right) } =\tan ^{ -1 }{ \left( \frac { 5 }{ 12 } \right) } \)
and \(\cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } =\tan ^{ -1 }{ \left( \frac { 4 }{ 3 } \right) } \)
\(R.H.S=\sin ^{ -1 }{ \left( \frac { 5 }{ 13 } \right) } +\cos ^{ -1 }{ \left( \frac { 3 }{ 5 } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 5 }{ 12 } \right) } +\tan ^{ -1 }{ \left( \frac { 4 }{ 3 } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { \frac { 5 }{ 12 } +\frac { 4 }{ 3 } }{ 1-\frac { 5 }{ 12 } \times \frac { 4 }{ 3 } } \right) } \)
\(\left[ \because \tan ^{ -1 }{ x } +\tan ^{ -1 }{ y } =\tan ^{ -1 }{ \left[ \frac { x+y }{ 1-xy } \right] } \right] \)
\(=\tan ^{ -1 }{ \left( \frac { \frac { 15+48 }{ 36 } }{ \frac { 36-20 }{ 36 } } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 63 }{ 16 } \right) } \)
\(=\tan ^{ -1 }{ \left( \frac { 63 }{ 65 } \right) } =R.H.S\)

\(\left( i \right) \ \begin{bmatrix} a & b \\ -b & a \end{bmatrix}+\begin{bmatrix} a & b \\ b & a \end{bmatrix}\)
\(=\begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix}=\begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}\)
\(\left( ii \right) \ \begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & { b }^{ 2 }+{ c }^{ 2 } \\ { a }^{ 2 }+{ c }^{ 2 } & { a }^{ 2 }+{ b }^{ 2 } \end{bmatrix}+\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}\)
\(=\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 }+2ab & { b }^{ 2 }+{ c }^{ 2 }+2bc \\ { a }^{ 2 }+{ c }^{ 2 }-2ac & { a }^{ 2 }+{ b }^{ 2 }-2ab \end{bmatrix}\)
\(=\begin{bmatrix} { \left( a+b \right) }^{ 2 } & { \left( b+c \right) }^{ 2 } \\ { \left( a-c \right) }^{ 2 } & { \left( a-b \right) }^{ 2 } \end{bmatrix}\)
\(\left( iii \right) \ \left[ \begin{matrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{matrix} \right] +\left[ \begin{matrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{matrix} \right]\)
\(=\left[ \begin{matrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{matrix} \right] \)
\(\left( iv \right) \quad \begin{bmatrix} \cos ^{ 2 }{ x } & \sin ^{ 2 }{ x } \\ \sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } \end{bmatrix}+\begin{bmatrix} \sin ^{ 2 }{ x } & \cos ^{ 2 }{ x } \\ \cos ^{ 2 }{ x } & \sin ^{ 2 }{ x } \end{bmatrix}.\)
\(=\begin{bmatrix} \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } & \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \\ \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } & \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \end{bmatrix}\)
\(=\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)

\(A={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
But in equilateral triangle of each side equal toa, area = \(\sqrt{3}a^2\over4\)
= \({\sqrt{3}a^2\over4}={1\over2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
\(\Rightarrow\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}={\sqrt{3}a^2\over2}\)
Hence,\(\left| \begin{matrix} x_1 &y_1 &1 \\x_2 &y_2 &1 \\x_3 &y_3 &1 \end{matrix} \right| ^2={3\over4}a^4\) is true.

k = 5

A = {1, 2, 3, 4, 5, 6}
R = {(x, y): y is divisible by x}
We know that any number (x) is divisible by itself.
⇒ (x, x) ∈R
∴ R is reflexive.
Now,
(2, 4) ∈R [as 4 is divisible by 2]
But,
(4, 2) ∉ R. [as 2 is not divisible by 4]
∴ R is not symmetric.
Let (x, y), (y, z) ∈ R. Then, y is divisible by x and z is divisible by y.
∴ z is divisible by x.
⇒ (x, z) ∈R
∴ R is transitive.
Hence, R is reflexive and transitive but not symmetric.

The given system of equation is
2x - y = 5 and x + y = 4.
The given system of equations can be written in the form of  A X = B, where 
\(A=\left[\begin{array}{cc} 2 & -1 \\ 1 & 1 \end{array}\right], X=\left[\begin{array}{l} x \\ z \end{array}\right] \text { and } B=\left[\begin{array}{l} 5 \\ 4 \end{array}\right]\)
\(\text { Now, }|A|=2(1)-(-1)(1)=2+1=3 \neq 0\)
∴ A is non-sinqular. 
Therefore,  A^-1 exists. 
Hence, the given system of equations is consistent. 

we have,\(A=\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]\) 
Here, \(|A|=\left|\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right|=14-12=2 \neq 0\) 
\(\therefore\) A^-1 exists.
Clearly,\(\operatorname{adj}(A)=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]\) 
\(\left[\text { if } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \text { , then adj }(A)=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]\right]\) 
\( \therefore A^{-1} =\frac{1}{|A|} \operatorname{adj}(A) \)
\(=\frac{1}{2}\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right] \) 
Now, consider RHS = 9 I - A
\(=9\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right] \)
\(=\left[\begin{array}{ll} 9 & 0 \\ 0 & 9 \end{array}\right]-\left[\begin{array}{cc} 2 & -3 \\ -4 & 7 \end{array}\right]=\left[\begin{array}{ll} 7 & 3 \\ 4 & 2 \end{array}\right]\) 
= 2A^-I [using Eq. (i)]
=LHS Hence proved

2 tan^-1(cos x) = tan^-1(2 cosec x)
\(\Rightarrow { tan }^{ -1 }\left( \frac { 2cosx }{ 1-{ cos }^{ 2 }x } \right) ={ tan }^{ -1 }(2\quad cosec\quad x)\)
\(\Rightarrow\) 2 cot x cosec x = 2 cosec x
\(\Rightarrow\) cot x = 1
\(\Rightarrow x={ cot }^{ -1 }(1)=\frac { \pi }{ 4 } \)

Any square matrix can be expressed
\(A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\)
we know that (A + A′) is a symmetric matrix and (A – A′) is a skew symmetric matrix. Since for any matrix A, (kA)′ = kA′, it follows that \(\frac{1}{2}\left(A+A^{\prime}\right)\) is symmetric matrix and \(\frac{1}{2}\left(A-A^{\prime}\right)\) is skew symmetric matrix. Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.

The set of all elements related to 1 is { a ∈ A : a = 1}.

(i) (b) \(\ln \triangle A B D\)
\(\tan \alpha  =\frac{B D}{A D}\)
\( =\frac{30}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\Rightarrow  \alpha  =30^{\circ}\)
\(\therefore  \sin \alpha =\sin 30^{\circ}=\frac{1}{2}\)
\(\Rightarrow  \alpha  =\sin ^{-1}\left(\frac{1}{2}\right)\)
(ii) (c) \(\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow \alpha =\cos ^{-1}\left(\frac{\sqrt{3}}{2})\right.\)
(iii) (d) \(\ln \triangle B DC\)
\(\tan \beta  =\frac{(BD)}{(C D)}\)
\(=\frac{30}{10 \sqrt{3}}=\sqrt{3} \)
\(\Rightarrow \beta =\tan ^{-1}(\sqrt{3})\)
(iv) \( \text { (c) Since, } \alpha=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \text {, } \)
\(\beta=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)
\(\therefore \quad \angle A B C=\pi-\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\)
\( =\pi-\frac{\pi}{2}=\frac{\pi}{2}\)
(v) (c) We know that domain and range of \(\cos ^{-1} x\) are [-1,1] and \([0, \pi]\) respectively.

(i) (d) Given, R = {(V₁,V₂) :  V₁,V₂ ∈ l}and both use their voting right in general election-2019. Since, X,Y∈ I⋅X exercised his voting right while Y did not cast her vote in general election-2019.
∴ Clearly, (X,Y)∉R
(ii) (a) Relation is symmetric.
∴(X,W)∈R
⇒(W,X)∈R
(iii) (a) Since, (F,F)∈R,F ∈ I and F use their voting right.
⇒R is reflexive.
⇒(F1,F2)∈R
⇒(F2,F1)∈R
R is symmetric.
and (F₁,F₂) ∈ R 
and  (F₂,F₃) ∈ R
⇒ (F₁,F₃) ∈ R 
(By transitive property)
(iv) (c) Given, relation RR is reflexive, symmetric and transitive.
∴R is equivalence relation.
(v) (a) Clearly, Mr. Shyam exercised his voting right in general election-2019, then Mr. Shyam is related to all those eligible voters who cast their votes.

Let the cost of 1 pen = Rs. x, the cost of 1 bag = Rs. y, and the cost of 1 instrument box = Rs. z
According to the question, we have
5x + 3y + z = 16, 2x + Y + 3z = 19, x + 2y + 4z = 25
This system of equation can be written as AX = B,
where \(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \) and \(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\) 
IAI = 5(4 - 6) - 3(8 - 3) + 1(4 - 1)
\(=-10-3(5)+3=-22 \neq 0\)
\(\therefore\)  A ^-1 exists.
Now, X = A^-1B, where \(A^{-1}=\frac{1}{|A|} \operatorname{adj} A\) 
Here, \(\operatorname{adj} A=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{\prime}=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad A^{-1}=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{c} 16 \\ 19 \\ 25 \end{array}\right]\) 
\(=\frac{1}{-22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right]=\frac{-1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]\) 
\(\therefore\) x = 1, y = 2, z = 5
Hence, cost of one pen, one bag and an instrument box isRs. 1, Rs. 2 and Rs. 5 respectively.
(i) (c) : Cost of one pen is Rs. 1.
(ii) (a) : Cost of one pen. and one bag = Rs. (1 + 2) = Rs. 3
(iii) (b) : Cost of one pen and one instrument box
= Rs. (1 + 5) = Rs. 6
(iv) (c) : According to the definition of determinant, determinartt is a number associated to a square matrix.
(v) (b) : Given matrix equation is AB = AC Pre-multiplying by A^-I on both sides, we get
\(A^{-1} A B=A^{-1} A C \Rightarrow\left(A^{-1} A\right) B=\left(A^{-1} A\right) C\) 
\(\Rightarrow \quad I B=I C \quad\left(\because A A^{-1}=A^{-1} A=I\right)\) 
\(\Rightarrow B=C\) 
Since A^-1 exists only if A i's non-singular
\(\therefore\) For B = C, A should be non-singular

(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.

(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion