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 \(x \geq 0, y \geq 0\) 

\(P=\left[ \begin{matrix} 2 & 3 \\ 1 & 0 \end{matrix} \right] \ Q=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right] \)

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

Given \({ a }_{ ij }=\frac { 1 }{ 3 } \left| -3i+j \right| \)
\(\therefore \ { a }_{ 23 }=\frac { 1 }{ 3 } \left| -6+3 \right| =1\)
 

Given \(A=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix};kA\begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \)
\({ A }^{ 2 }=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}=\begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix}\) [multiplying row by column]
\({ A }^{ 2 }=kA\Rightarrow \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \)
\(\Rightarrow k=2A\)
On comparing with Eq. (ii), we got k = 2

1

Given that,
Minimise and Maximise Z = 3x + 9y       ...(i)
Subject to the constraints are
\(\begin{aligned} x+3 y & \leq 60 \end{aligned}\)        ...(ii)
\(\begin{aligned} x+y & \geq 10 \end{aligned}\)          ...(iii)
\(\begin{aligned} x & \leq y \end{aligned}\)                  ...(iv)
\(\begin{aligned} x \geq 0, y & \geq 0 \end{aligned}\)         ...(v)

First of all,let us plot the graph of the feasible region of the system of linear inequalities (ii) to (v). The feasible region ABCDA is shown in the figure.
Note That the region is bounded. The coordinates of the corner points A, B, C and D are (0, 10), (5, 5), (15, 15) and (0, 20), respectively.

We, now find the minimum and maximum value of Z. From the table, we find that the minimum value of Z is 60 at the point B(5, 5) of the feasible region.
The maximum value of Z on the feasible region occurs at the two corner points C (15, 15) and D (0, 20) and it is 180 in each case.
Remark Observe that in the above example, the problem has multiple optimal solutions at the corner points C and D, i.e. the both points produce same maximum value 180.
In such cases, you can see that every point on the line segment CD joining the two corner points C and D also give the same maximum value. Same is also true in the case, if the two points produce same minimum value.

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).

Given, \(A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\) 
Now, \(A^{2}=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\) 
\(=\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right]=\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\) 
and 
\(A^{3} =A \cdot A^{2}=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right] \)
\(=\left[\begin{array}{ccc} 5+0+16 & 0+0+0 & 8+0+26 \\ 0+4+8 & 0+8+0 & 0+10+13 \\ 10+0+24 & 0+0+0 & 16+0+39 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right] \) 
 Since, A³ - 6A² + 7A + kI³ = 0
\(\begin{aligned} \therefore &\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] &+k\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{aligned}\) 
\( \Rightarrow\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right] +\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{lll} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \) 
 \(\Rightarrow\left[\begin{array}{ccc} 21-30+7+k & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+k & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+k \end{array}\right]\) \(=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) 
 \(\Rightarrow\left[\begin{array}{ccc} -2+\mathrm{A} & 0 & 0 \\ 0 & -2+k & 0 \\ 0 & 0 & -2+k \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) 
On equating the corresponding elements, we get
-2 + k = 0
k = 2

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.

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.

The given system of equations is:
5x + 2y = 4
7x + 3y = 5
These can be written as AX=B
=X = A^-1B ....(1)
where \(A=\begin{bmatrix} 5&2\\7&3\end{bmatrix},X=\begin{bmatrix}x\\y \end{bmatrix}and\ B=\begin{bmatrix}4\\5 \end{bmatrix}\)
\(|A|=\begin{vmatrix} 5&2\\ 7&3\end{vmatrix}=15-14=1\ne0\Rightarrow A^{-1}\) exists.
Now \(adj\ A=\begin{bmatrix}3&-7\\-2&5 \end{bmatrix}'=\begin{bmatrix} 3&-2\\-7&5\end{bmatrix}\)
\(A^{-1}={1\over |A|}(adj\ A)={1\over1}\begin{bmatrix}3&-2\\-7&5 \end{bmatrix}.\)
From (1), \(X=\begin{bmatrix}3&-2\\-7&5 \end{bmatrix}\begin{bmatrix} 4\\5\end{bmatrix}\)
\(=\begin{bmatrix} 12-10\\-28+25\end{bmatrix}\)
\(\begin{bmatrix} x\\y\end{bmatrix}=\begin{bmatrix}2\\-3 \end{bmatrix}\)
Hence x = 2, y = -3

(i) Area of the triangle
\(={1\over2}\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
\({1\over2}\begin{vmatrix}k&0&1\\4&0&1\\0&2&1 \end{vmatrix}\)
\({1\over2}(-2)[k-4]\)
\(-(k-4)\)
By the equation, -(k - 4) = \(\pm4\)
Taking +ve sign, -(k - 4) = 4 \(\Rightarrow\) k = 0
Taking -ve sign, -(k - 4) = -4 \(\Rightarrow\) k = 8
Hence k = 0, 8
(ii)Area ofg the triangle \(={1\over2}\begin{vmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1 \end{vmatrix}\)
\({1\over2}\begin{vmatrix}-2&0&1\\0&4&1\\0&k&1 \end{vmatrix}\)
\({1\over2}(-2)[k-4]\)
\(-(k-4)\)
By the equation, k - 4 = \(\pm4\)
Taking +ve sign, k - 4 = 4 \(\Rightarrow\) k = 0
Taking -ve sign, k - 4 = -4 \(\Rightarrow\) k = 8
Hence k = 0, 8

We have \(A^2=A \cdot A=\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]\)
Hence \(\mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\left[\begin{array}{cc} 7 & 12 \\ 4 & 7 \end{array}\right]-\left[\begin{array}{cc} 8 & 12 \\ 4 & 8 \end{array}\right]+\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=\mathrm{O}\)
\(\begin{array}{l} \mathrm{A}^2-4 \mathrm{~A}+\mathrm{I}=\mathrm{O} \\ \mathrm{A} \mathrm{A}-4 \mathrm{~A}=-\mathrm{I} \end{array}\)
or \( \mathrm{A} \mathrm{A}\left(\mathrm{A}^{-1}\right)-4 \mathrm{AA}^{-1}=-\mathrm{IA}^{-1} \) (Post multiplying by A–1 because |A| ≠ 0)
or \( {A}\left(\mathrm{~A} \mathrm{~A}^{-1}\right)-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{AI}-4 \mathrm{I}=-\mathrm{A}^{-1} \)
or \(\mathrm{A}^{-1}=4 \mathrm{I}-\mathrm{A}=\left[\begin{array}{ll} 4 & 0 \\ 0 & 4 \end{array}\right]-\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)
Hence \(A^{-1}=\left[\begin{array}{cc} 2 & -3 \\ -1 & 2 \end{array}\right]\)

\( \Delta=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ =a_{11} M_{11}-a_{12} M_{12}+a_{13} M_{13} \\ =1 (-40)-3(-10)+(-2)(35) \\ =-40+30-70=-80 \)

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

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).

As \(\begin{vmatrix} 2x & -1 \\ 4 & 2 \end{vmatrix}=\begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix}\)
⇒ 4x + 4 = 3 - 0
⇒ x = \(-\frac { 1 }{ 4 } \)

As in skew symmetric matrix, a_ij = -a_ji
⇒ a_ii = – a_ii
⇒ 2a_ii = 0
⇒ a_ii = 0, i.e. diagonal elements are zeroes.

By definition.

As a₂₃ = \(\frac { { (-2+6) }^{ 2 } }{ 5 } =\frac { 16 }{ 5 } \)

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):

Let A be the 2 x 3 matrix representing the annual sales of products in two markets.

Let B be the column matrix representing the sale price of each unit of products x, y, z.
\(\therefore \quad B=\left[\begin{array}{c} 2.5 \\ 1.5 \\ 1 \end{array}\right]\) 
Now, revenue = sale price x number of items sold
\(=\left[\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}\right]\left[\begin{array}{c} 2.5 \\ 1.5 \\ 1 \end{array}\right]\) 
\(=\left[\begin{array}{l} 25000+3000+18000 \\ 15000+30000+8000 \end{array}\right]=\left[\begin{array}{l} 46000 \\ 53000 \end{array}\right]\) 
Therefore, the revenue collected from Market I = Rs.46000 and the revenue collected from Market II = Rs. 53000.
(i) (c)
(ii) (b)
(iii) (d) : Let C be the column matrix representing cost price of each unit of products x, y, z.
Then, \(C=\left[\begin{array}{c} 2 \\ 1 \\ 0.5 \end{array}\right]\) 
\(\therefore\) Total cost in each market is given by 
\(A C=\left[\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}\right]\left[\begin{array}{c} 2 \\ 1 \\ 0.5 \end{array}\right]\) 
\(=\left[\begin{array}{c} 20000+2000+9000 \\ 12000+20000+4000 \end{array}\right]=\left[\begin{array}{l} 31000 \\ 36000 \end{array}\right]\) 
Now, Profit matrix = Revenue matrix, Cost matrix
\(=\left[\begin{array}{l} 46000 \\ 53000 \end{array}\right]-\left[\begin{array}{l} 31000 \\ 36000 \end{array}\right]=\left[\begin{array}{l} 15000 \\ 17000 \end{array}\right]\) 
Therefore, the gross profit from both the markets
= Rs.15000+ Rs.17000= Rs.32000
(iv) (a) : We have \(A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\) 
\(\therefore \quad A^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I\) 
(v) (a) : We have, (AB' -BA')' = (B')'A' - (A')'B'
= BA' - AB' = -(AB' - BA')
Thus, AB' - BA' is a skew-symmetric matrix.

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

(c) Assertion is correct, reason is incorrect