Our problem is to maximise
Z = 3x + 4y                 ...(i)
Subject to the constraints are
\(\begin{aligned} x+y & \leq 4 \end{aligned}\)           ...(ii)
\(\begin{aligned} x & \geq 0, y \geq 0 \end{aligned}\)       ...(iii)
Table for the line x + y = 4 is

On putting (0, 0) in the inequality x+ y \(\leq\) 4, we have
0 + 0 \(\leq\) 4
\(\Rightarrow\)    0 \(\leq\) 4 (which is true)
So, the half plane is towards the origin

So, the feasible region lies in the Ist quadrant.
\(\therefore\) Feasible region is OABO.
The corner points of the feasible region are O(0, 0), A(4, 0) and B(0, 4). The values of Z at these points are as follows

Therefore, the maximum value of Z is 16 at the point B(0, 4).

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).

Expanding along first row, we get
\(\Delta\)= 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0

\(y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\}, \)
\(\text { Let } \text x=\sin \theta, \)
\(\text { then } y=\sin ^{-1}\left\{\frac{5 \sin \theta+12 \cos \theta}{13}\right\} \)
\(=\sin ^{-1}\left\{\frac{5}{13} \sin \theta+\frac{12}{13} \cos \theta\right\} \)
\(\text { Let } \frac{5}{13}=\cos \alpha, \frac{12}{13}=\sin \alpha, \)
\(\text { We notice } \sin ^{2} \alpha+\cos ^{2} \alpha=1 .\)
\(\Rightarrow y =\sin ^{-1}\{\sin \theta \cos \alpha+\cos \theta \sin \alpha\} \)
\(=\sin ^{-1}\{\sin (\theta+\alpha)\}=\theta+\alpha \)
\(=\sin ^{-1} x+\cos ^{-1} \frac{5}{13} \)
\(\Rightarrow \frac{d y}{d x} =\frac{1}{\sqrt{1-x^{2}}}\)
\(=\frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } \)

From (i) , (ii) and (iii), we note that function is continuous at x=\(\frac { 1 }{ 2 } \).

\(\text { Given }\left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0, a \neq 0\)
\(\Rightarrow\left|\begin{array}{rrc} a & 0 & x \\ -a & a & x \\ 0 & -a & x+a \end{array}\right|=0\)
\(\left[\text { by performing } C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\right]\)
\(\Rightarrow\left|\begin{array}{ccc} a & 0 & x \\ 0 & a & 2 x \\ 0 & -a & x+a \end{array}\right|=0\left[\text { by performing } R_{2} \rightarrow R_{2}+R_{1}\right]\)
\(\Rightarrow a\left(a x+a^{2}+2 a x\right)=0 \text { [on expanding along } \left.C_{1}\right] \)
\(\Rightarrow 3 a x+a^{2}=0 \Rightarrow x=-\frac{a}{3}, a \neq 0\)

Find AB, A', B', B'A', compare.Proceed.

\(A^{2}-4 A-5 I\)
\(=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{array}\right]\)
\(-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{array}\right]-\left[\begin{array}{lll} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{array}\right]-\left[\begin{array}{lll} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]\)
\(=\left[\begin{array}{lll} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{array}\right]\)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\text { Hence, } A^{2}-4 A-5 I=O\)