Let x necklaces and y bracelets manufactured by small firm.
\(\begin{array}{c|c|c|c} \hline \text { Item } & \text { Number } & \begin{array}{c} \text { Manufactures } \\ \text { times (in hours) } \end{array} & \text { Profit (in Rs.) } \\ \hline \text { Necklaces } & x & x / 2 & 100 x \\ \text { Bracelets } & y & y & 300 y \\ \text { Total } & x+y & \frac{x}{2}+y & 100 x+300 y \\ \hline \text { Availability } & 24 & 16 & \\ \hline \end{array}\) 
Our problem is to maximise Z = 100x + 300y subject to constraints are 
\(x \geq 1, y \geq 1 \)
\(x+y \leq 24 ; \frac{1}{2} x+y \leq 16\)

Suppose tailor A work for x days and tailor B work for y days.
The given data can be written in the tabular form as follows
\(\begin{array}{c|c|c|c} \hline \text { Tailor } & \begin{array}{c} \text { Number } \\ \text { of shirts } \end{array} & \begin{array}{c} \text { Number of } \\ \text { trousers } \end{array} & \text { Cost/day } \\ \hline A & 6 & 4 & Rs. 300 \\ B & 10 & 4 & \text { Rs. } 400 \\ \hline \text { Minimum requirement } & 60 & 32 & \\ \hline \end{array}\) 
Required linear programming problem is
Min (Z) = 300x + 400y
subject to constraints
\(6 x+10 y \geq 60\) 
\(4 x+4 y \geq 32\)  and \(x \geq 0, y \geq 0\) 

\(AB=\left[\begin{array}{lll}39 & 57 & 21 \\ 24 & 34 & 14\end{array}\right]\) 'BA does not exist

1 x 1, 3 x 3

\(A=\left[ \begin{matrix} 0 & x & -4 \\ -2 & 0 & -1 \\ y & -1 & 0 \end{matrix} \right] \)
\({ A }^{ \prime }=\left[ \begin{matrix} 0 & -2 & y \\ x & 3 & -1 \\ -4 & -1 & 0 \end{matrix} \right] \)
For skew symmetric
\(A={ -A }^{ \prime }\)
\(\Rightarrow \left[ \begin{matrix} 0 & x & -4 \\ -2 & 0 & -1 \\ y & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -2 & y \\ x & 3 & -1 \\ -4 & -1 & 0 \end{matrix} \right] \)
x = 2, y = 4

\(\text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right], \text { then adj } A=\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \text { . }\)
So, Adj \(A=\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}\)

1

We have, maximise, Z = 60x + 40y             ...(i)
Subject to the constraints, x + 2y \(\leq 12\)     ...(ii)
\(\begin{aligned} 2 x+y & \leq 12 \end{aligned}\)                                   ...(iii)
\(\begin{aligned} 4 x+5 y & \geq 20 \end{aligned}\)                                 ...(iv)
and \(\begin{aligned} x, y & \geq 0 \end{aligned}\)                                     ...(v)
Table for  line x + 2y = 12 is

So, the line x + 2y = 12 is passing through the points (0, 6) and (12, 0).
On putting (0,0) in the inequality \(x+2 y \leq 12\) we get \(0+2(0) \leq 12 \Rightarrow 0 \leq 12\), which is true
So,the half plane is towards the origin.
Table for line 2x + y = 12 is 

So, the line 2x + y = 12 is passing through the points (0, 12) and (6, 0).
On putting (0, 0) in the inequality \(2 x+y \leq 12\) we get
\(2(0)+0 \leq 12 \Rightarrow 0 \leq 12\) , which is true.
So, the half plane is towards the origin.
Table for line 4x + 5y = 20 is

So, the line 4x + 5y = 20 is passing through the points (0,4) and (5, 0).
On putting (0, 0) in the inequality \(4 x+5 y \geq 20\), we get 4(0) + 5(0) \(\geq\) 20 \(\Rightarrow\) 0 > 20 which is not true.
So, the half plane is away from the origin.
Also, x, y \(\geq\) 0
So, the region lies in Ist quadrant.

On solving Eqs. x + 2y = 12 and 2x + y = 12, we get D(4, 4)
Clearly, the feasible region is ABCDEA.
The corner points of the feasible region are A(0, 4), B(5, 0), C(6, 0), D(4, 4) and E(0, 6).
The value of Z at corner points are given below.

The maximum value of Z is 400 at D(4, 4).

Let x and y units of packet of mixes are purchased from 5 and T respectively. If Z is the total cost, then
Z = 10x + 4y ...(i)
is objective function which we have to minimize
Here constraints are:
4x + y \(\ge\)80..(ii)
2x + y \(\ge\)60.(iii) 
Also, x \(\ge\) 0...(iv)
y \(\ge\)0
On plotting the graph of above constraints or inequalities (ii), (iii), (iv) and (v),
we get shaded region having corner point A, P, B as feasible region. For co-ordinate of P

Point of intersection of
2x + y = 60...(vi)
and 4x + y = 80....(vii)
From (vi) - (vii),
2x + y - 4x - y = 60 - 80
\(\Rightarrow\) -2x = -20
\(\Rightarrow\) x = 10
\(\Rightarrow\) y = 40
\(\because\) Co-ordinate of P = (10, 40)
Now the value of Z is evaluated at corner point the following table:

Since feasible region is unbounded. Therefore we have to draw the graph of the inequality.
10x + 4y < 260 ...(viii)
Since the graph of inequality (viii) does not have any point common.
So the minimum value of Z is 260 at (10, 40). i.e., minimum cost of each bottle is ~ 260 if the company purchases 10 packets of mixes from 5 and 40 packets of mixes from supplier T.

We have :  \(B=\left[ \begin{matrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{matrix} \right] \)
\(\therefore \ B'=\left[ \begin{matrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{matrix} \right] .\)
Let \(\mathrm{P}=\frac{1}{2}\left(\mathrm{~B}+\mathrm{B}^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rrr} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{array}\right]=\left[\begin{array}{ccc} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \frac{-3}{2} & 3 & 1 \\ \frac{-3}{2} & 1 & -3 \end{array}\right]\) 
Now, \(\mathrm{P}^{\prime}=\left[\begin{array}{ccc} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \frac{-3}{2} & 3 & 1 \\ \frac{-3}{2} & 1 & -3 \end{array}\right]=\mathrm{P}\)
Thus, \(P=\frac{1}{2}\left(B+B^{\prime}\right)\) which is symmetric.
Also let, \(Q=\frac{1}{2}\left(B-B^{\prime}\right)=\frac{1}{2}\left[\begin{array}{rrr} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{array}\right]\)
Then, \(\mathrm{Q}^{\prime}=\left[\begin{array}{ccc} 0 & \frac{1}{2} & \frac{5}{3} \\ \frac{-1}{2} & 0 & -3 \\ \frac{-5}{2} & 3 & 0 \end{array}\right]=-\mathrm{Q}\)
Thus \(\mathrm{Q}=\frac{1}{2}\left(\mathrm{~B}-\mathrm{B}^{\prime}\right)\) is a skew-symmetric.    
\(\mathrm{P}+\mathrm{Q}=\left[\begin{array}{ccc} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \frac{-3}{2} & 3 & 1 \\ \frac{-3}{2} & 1 & -3 \end{array}\right]+\left[\begin{array}{ccc} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{array}\right]=\left[\begin{array}{rrr} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{array}\right]=\mathrm{B}\)
Thus, B is represented as the sum of a symmetric and a skew symmetric matrix.

Since A and B are symmetric matrices, [Given]
\(\therefore\ A\prime =A\ and\ B\prime =B...(1)\)
\((i)\ (A+B)\prime =A\prime +B\prime =A+B.\ \)    [Using (1)]
Hence, \(A+B\) is a symmetric matrix 
\((ii)\ (AB-BA)\prime =(AB)\prime -(BA)\prime \)
\(=B\prime A\prime -A\prime B\prime =BA-AB\)    [Using (1)]
\(=-(AB-BA).\)
Hence, \(AB-BA\) is a skew-symmetric matrix.
\((iii)\ (AB+BA)\prime =(AB)\prime +(BA)\prime \)
\(=B\prime A\prime +A\prime B\prime =BA+AB\quad \)  [Using (1)]
\(=AB+BA.\)
  Hence, \(AB+BA\) is a skew-symmetric matrix.

\( \left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right| \)
\( =x\left(-x^2-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta) \)
\( =-x^3-x+x \sin ^2 \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^2 \theta \)
\( =-x^3-x+x\left(\sin ^2 \theta+\cos ^2 \theta\right) \)
\(=-x^3-x+x \)
\(=-x^3 \text { (Which is Independent of } \theta \text { ) }\)

Consider the product \(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]\)
\(=\left[\begin{array}{ccc} -2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8 \end{array}\right]=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\)
Hence, \(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1}=\left[\begin{array}{ccc} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{array}\right]\)
Now, given system of equations can be written, in matrix form, as follows
\(\left[\begin{array}{ccc} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]\)
\(\begin{aligned} &x \\ &y \\ &z \end{aligned}=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{array}\right]^{-1}\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]=\left[\begin{array}{lll} 2 & 0 & 1 \\ 9 & 2 & 3 \\ 6 & 1 & 2 \end{array}\right]\left[\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right]\)
\(=\left[\begin{array}{r} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{array}\right]=\left[\begin{array}{l} 0 \\ 5 \\ 3 \end{array}\right]\)
Hence, x = 0, y = 5, z = 3

\(\text { Let } A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{array}\right]\)
\(\text { We have, }|A|=1(10-0)-2(0-0)+3(0-0)=10\)
\(\text { Now, }A_{11}=10-0=10, A_{12}=-(0-0)=0, A_{13}=0-0=0\)
\(A_{21}=-(10-0)=-10, A_{22}=5-0=5, A_{23}=-(0-0)=0\)
\(A_{31}=8-6=2, A_{32}=-(4-0)=-4, A_{33}=2-0=2 \)
\(\therefore \operatorname{adj} A=\left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right] \)
\(\therefore A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{10}\left[\begin{array}{ccc} 10 & -10 & 2 \\ 0 & 5 & -4 \\ 0 & 0 & 2 \end{array}\right]\)

Using equality of matrices
3a + 4b = 2, c + d = 5, 2c - d = - 5, a - 2b = 4
Solving for a, b, c and d, we get
a = 2, b = -1, c = 0,d = 5

(i) Our problem is to minirnise Z = 3x + 5y 
subject to constraints \(x+3 y \geq 3\) 
\(x+y \geq 2\) 
and \(x \geq 0, y \geq 0\) .
(ii) Table for line x + 3y = 3 is 
\(\begin{array}{c|c|c} \hline x & 0 & 3 \\ \hline y & 1 & 0 \\ \hline \end{array}\) 
So, the line passes through the points (0, 1) and (3, 0).
On putting (0, 0) in the inequality \(x+3 y \geq 3\), we get
\(0+3 \times 0 \geq 3\) 
\(\Rightarrow 0 \geq 3\), which is not true
So, the half plane is away from the origin.
Also, \(x, y \geq 0\) so the feasible region lies in the I quadrant.
Table for line x + y = 2 is
\(\begin{array}{|c|c|c|} \hline x & 0 & 2 \\ \hline y & 2 & 0 \\ \hline \end{array}\) 
So, the line passes through the points (0, 2) and (2, 0). On putting (0, 0) in the inequality \(x+y \geq 2\), we get \(0+0 \geq 2 \Rightarrow 0 \geq 2\), which is not true

So, the half plane is away from the origin. It can be seen that the feasible region is unbounded. On solWng equations x + y = 2 and x + 3y = 3, we get
\(x=\frac{3}{2} \text { and } y=\frac{1}{2}\) 
\(\therefore \text { Intersection point is } B\left(\frac{3}{2}, \frac{1}{2}\right)\) 
The corner points of the feasible region are A(3,0), \(B\left(\frac{3}{2}, \frac{1}{2}\right) \text { and } C(0,2)\) 
The values of Z at the corner points are given below
\(\begin{array}{c|c} \hline \text { Corner points } & z=3 x+5 y \\ \hline A(3,0) & z=3 \times 3+5 \times 0=9 \\ B\left(\frac{3}{2}, \frac{1}{2}\right) & z=3 \times \frac{3}{2}+5 \times \frac{1}{2}=7 \\ C(0,2) & z=3 \times 0+5 \times 2=10 \\ \hline \end{array}\) 
As the feasible region is unbounded, therefore 7 may or may not be the minimum value of Z.
For this, we draw the graph of the inequality \(3 x+5 y<7\) and check whether the resulting half plane has points in common with the feasible region or not. It can be seen that, the feasible region has no common point with \(3 x+5 y<7\) 
Therefore, the minimum value of Z is 7 at \(B\left(\frac{3}{2}, \frac{1}{2}\right)\).

The system of constraints is:
\(x+y\le 4\) ..(1)
and \(x\ge 0,y\ge 0.\) ...(2)
The shaded region in the following figure is the feasible region determined by the system of constraints (1)-(2)
It is observed that the feasible region OAB is bounded.
Thus we use Corner Point Method to determine the maximum value of Z, where:
Z = 3x + 4y...(3)

The co-ordinates of O, A and B are (0, 0), (4, 0) and (0, 4) respectively.
We evaluate Z at each corner point.

Hence, \(Z_{ max }=16\) at the point (0, 4)

We have   \(=\left[\begin{array}{c} 4000+50000+250000 \\ 120000+100000+500000 \end{array}\right]=\left[\begin{array}{l} 304,000 \\ 720,000 \end{array}\right] \begin{aligned} &\rightarrow X \\ &\rightarrow Y \end{aligned}\)
\(=\left[\begin{array}{ll} 340,000 \\ 720,000 \end{array}\right] \begin{aligned} &\rightarrow \mathrm{X} \\ &\rightarrow \mathrm{Y} \end{aligned}\)
So the total amount spent by the group in the two cities is Rs. 340,000 paise and Rs. 720,000 paise, i.e., Rs. 3400 and Rs. 7200, respectively.

(i) Let (x, y) be the third point on the joining (1, 2) and (3, 6)
The area of triangle having vertices (x, y), (1, 2) and (3, 6)
\(={1\over2}\begin{vmatrix} x&y&1\\1&2&1\\3&6&1\end{vmatrix}\)
Since the three points are collinear,
=\(\begin{vmatrix} x&y&1\\1&2&1\\3&6&1 \end{vmatrix}=0\)
= x(2-6) - y(1-3) + 1(6-6)
= -4x + 2y = 0
= 2x - y = 0
(ii) As in part (i)
Area \({1\over2}\begin{vmatrix} x&y&1\\3&1&1\\9&3&1\end{vmatrix}=0\Rightarrow\begin{vmatrix} x&y&1\\3&1&1\\9&3&1\end{vmatrix}=0\)
= x(1 - 3) - y(3 - 9) + 1(9 - 9) = 0
= -2x + 6y = 0
= x - 3y = 0

\(We\quad have:\quad 2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
\(\Rightarrow \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\( \Rightarrow \begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix} y+2 & 6 \\ 1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}.\)
Equating corresponding elements:
2x + 2 = 8 and y + 2 = 5
\(\Rightarrow \) 2x = 6 and y = 3.
Hence, x = 3 and y = 3.

We get \(\begin{vmatrix}1&a&bc\\0&b-a&c(a-b)\\0&c-a&b(a-c) \end{vmatrix}\)
Taking factors (b – a) and (c – a) common from R₂ and R₃, respectively, we get
\((b-a)(c-a)\begin{vmatrix} 1&a&bc\\0&1&-c\\0&1&-b\end{vmatrix}\)
= (b – a) (c – a) [(– b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)

\(\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=\pm 35\)

Area of triangle, \(\Delta=\frac{1}{2}\left|\begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|\)

Given \(\left|\begin{array}{cc} x & 2 \\ 18 & x \end{array}\right|=\left|\begin{array}{cc} 6 & 2 \\ 18 & 6 \end{array}\right| \Rightarrow x^{2}-36=36-36\) 
\(\Rightarrow \quad x^{2}=36 \Rightarrow x=\pm 6\)

We know that if a matrix is of order m x n, then it has mn elements. Thus, to find all possible orders of a matrix with 8 elements, we will find all ordered pairs of natural numbers, whose product is 8. Thus, all possible ordered pair are (1,8), (8, I), (2, 4), (4, 2).

According to the question, when length is decreased by 50 m and breadth is increased by 50 m
(x-50)(y+50) =xy
x - y = 50
and when length is decreased by 10 m and breadth is decreased by 20 m .
(i) (b) x-y=50 and 2 x+y=550
(ii) (a) Eqs. (i) and (ii) can be written in matrix form as
\(\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 50 \\ 550 \end{array}\right]\)
(iii) (c) We have, \( {\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{c} 50 \\ 550 \end{array}\right] } \)
\(\Rightarrow  {\left[\begin{array}{c} x \\ y \end{array}\right]=\left[\begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array}\right]^{-1}\left[\begin{array}{c} 50 \\ 550 \end{array}\right] }\)
\(\therefore \quad(x-10)(y-20)=x y-5300\)
\( \Rightarrow \quad 2 x+y=550\)
\( =\frac{1}{1-(2)(-1)}\left[\begin{array}{cc} 1 & 1 \\ -2 & 1 \end{array}\right]\left[\begin{array}{c} 50 \\ 550 \end{array}\right] \)
\(=\frac{1}{3}\left[\begin{array}{cc} 50+550 \\ c & b \\ -100+550 \end{array}\right]=\frac{1}{3}\left[\begin{array}{l} 600 \\ 450 \end{array}\right]=\left[\begin{array}{l} 200 \\ 150 \end{array}\right]\)
\therefore x = 200 and y=150
Lenght of rectangular field
\(\Rightarrow x=200 \mathrm{~m}\)
(iv) (a) Breadth of rectangular field \(y=150 \mathrm{~m}\)
(v) (b) Area of rectangular field =200 x 150 \(=30000 \mathrm{~m}^2\)

(i) (b): Let B(x, y) be the point of intersection of the given lines
2x + 5y = 100 ....(i)
and \(\frac{x}{25}+\frac{y}{40}=1 \Rightarrow 8 x+5 y=20\)...(ii)
Solving (i) and (ii), we get
\(x=\frac{50}{3}, y=\frac{40}{3}\)
ஃ The point of intersection \(B(x, y)=\left(\frac{50}{3}, \frac{40}{3}\right)\)
(ii) (d): The corner points of the feasible region shown in the given graph are
\((0,0), A(25,0), B\left(\frac{50}{3}, \frac{40}{3}\right), C(0,20)\)
(iii) (a): Here Z = x + y

Thus, max Z = 30 occurs at point \(\left(\frac{50}{3}, \frac{40}{3}\right)\)
(iv) (b): 

(v) (c):

(i) (c) : Let Rs. A, Rs. B and Rs.C be the cost incurred by the organisation for villages X, Y and Z respectively. Then A, B, C will be given by the following matrix equation.
\(\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\left[\begin{array}{l} 50 \\ 20 \\ 40 \end{array}\right]=\left[\begin{array}{c} A \\ B \\ C \end{array}\right]\) 
\(\Rightarrow\left[\begin{array}{l} A \\ B \\ C \end{array}\right]=\left[\begin{array}{c} 400 \times 50+300 \times 20+100 \times 40 \\ 300 \times 50+250 \times 20+75 \times 40 \\ 500 \times 50+400 \times 20+150 \times 40 \end{array}\right]\) 
(ii) (c)
(iii) (b)
(iv) (c) : Total number of toilets that can be expected in each village is given by the following matrix.
\(\begin{array}{l} X \\ Y \\ Z \end{array}\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\)\(\left[\begin{array}{c} 2 / 100 \\ 4 / 100 \\ 20 / 100 \end{array}\right]\) 
\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 8+12+20 \\ 6+10+15 \\ 10+16+30 \end{array}\right]\)=\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 40 \\ 31 \\ 56 \end{array}\right]\) 
(v) (d)

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

(c) Assertion is correct, reason is incorrect