\(A=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right]\)
\(\therefore \ { A }^{ -1 }=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix} \right]\)
\(Now\ { A }^{ -1 }A=\left[ \begin{matrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 3-2+0 & 6+6-12 & -6+0+6 \\ 1-1+0 & 2+3-4 & -2+0+2 \\ 2-2+0 & 4+6-10 & -4+0+5 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =I,\text{ Which is true.}\)

Here a*1 = a mod 6 = a
and 1*a = a mod 6 = \(a\forall a\in A\).
Thus 1 is the identity operation for '*'.
and 5*5 = 25 mod 6 = 1
\(\Rightarrow \)1 is the inverse of 1 and 5 is the inverse of 5.
Also, \(\forall a\in A\), other than 1 and 5, we cannot find \(b\) such that a*b = ab mod 6 = 1

\(\text {Consider } \tan ^{-1}\left[2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right] \)
\(=\tan ^{-1}\left[2 \sin \left(2 \cdot \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2 \sin \frac{\pi}{3}\right] \)
\(=\tan ^{-1}\left(2 \cdot \frac{\sqrt{3}}{2}\right)=\tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \)

A matrix A is said to be symmetric if A^T = A,
\(A+A^{T}=\left[\begin{array}{ll} 4 & 1 \\ 5 & 8 \end{array}\right]+\left[\begin{array}{ll} 4 & 5 \\ 1 & 8 \end{array}\right]=\left[\begin{array}{lc} 8 & 6 \\ 6 & 16 \end{array}\right] \)
\(\left(A+A^{T}\right)^{T}=\left[\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right]^{T}=\left[\begin{array}{cc} 8 & 6 \\ 6 & 16 \end{array}\right]\)
=(A + A^T)^T = A + A^T

Use A=lA. Proceed. [Refer 2]

\(\left[ \begin{matrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{matrix} \right] \)

\( \Delta=\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right| \)
Expanding along \(\mathrm{C}_3\),
we have:
\( \Delta =-\sin \alpha\left(-\sin \alpha \sin ^2 \beta-\cos ^2 \beta \sin \alpha\right)+\cos \alpha\left(\cos \alpha \cos ^2 \beta+\cos \alpha \sin ^2 \beta\right) \)
\(=\sin ^2 \alpha\left(\sin ^2 \beta+\cos ^2 \beta\right)+\cos ^2 \alpha\left(\cos ^2 \beta+\sin ^2 \beta\right) \)
\(=\sin ^2 \alpha(1)+\cos ^2 \alpha(1)  =1 \)

Let X = denote the random variable. X = 0, 1, 2n = 2, p = 1/4, q = 3/4

Mean = \(\sum { { x }_{ i }{ p }_{ i }=\frac { 1 }{ 2 } } \)
Variance = \(\sum { { x }_{ i }^{ 2 }{ p }_{ i } } -{ \sum { ({ x }_{ i }{ p }_{ i } } ) }^{ 2 }\)
= \(\frac { 5 }{ 8 } -\frac { 1 }{ 4 } =\frac { 3 }{ 8 } \)

Let, \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 12 } =\lambda\)
So, co-ordinates of any random point on this line, M
As the line intersect the plane, x-y+z-5 = 0
So, \( \left( 3\lambda +2 \right) -\left( 4\lambda -1 \right) +\left( 2\lambda +2 \right) -5=0\)
\(\Rightarrow\)  \(3\lambda +2-4\lambda +1+2\lambda +2-5=0\)
\(\Rightarrow\)  \(\lambda =0\)
So, required co-ordinates of point of intersection M(2, -1, 2)
Now the vector parallel to the given line,\(\vec { b } =3\hat { i } +4\hat { j } +2\hat { k } \) and the normal vector to the plane  \(\vec { m } =\hat { i } -\hat { j } +\hat { k } \).
Let \(\theta \) be the angle between the line and the plane, then by formula \(\vec { b } .\vec { m } =\left| \vec { b } \right| \left| \vec { m } \right| \left| sin\ \theta \right| \)
or  \(sin\ \theta =\frac { \vec { b } .\vec { m } }{ \left| \vec { b } \right| \left| \vec { m } \right| } \)
\(=\frac { \left( \hat { 3i } +4\hat { j } +2\hat { k } \right) .\left( \hat { i } -\hat { j } +\hat { k } \right) }{ \sqrt { { \left( 3 \right) }^{ 2 }+{ \left( 4 \right) }^{ 2 }+{ \left( 2 \right) }^{ 2 } } .\sqrt { { \left( 1 \right) }^{ 2 }+{ \left( -1 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 }} } =\frac { 3-4+2 }{ \sqrt { 87 } } =\frac { 1 }{ \sqrt { 87 } } \)
or   \(\theta ={ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 87 } } \right) \)

Let the 1^st paycheck be x (integer).
Mrs. Rodger got a weekly raise of 145.
So after completing the 1st week she will get (x+145).
Similarly after completing the 2^nd week she will get (x +145) + 145.
= (x + 145 + 145)
= (x + 290)
So in this way end of every week her salary will increase by 145.

 line is  \(\vec { r } .(\hat { i } -2\hat { j } +3\hat { k } )+\lambda (-3\hat { i } +5\hat { j } +4\hat { k } )\)
In Cartesian from : \(\frac { x-1 }{ -3 } =\frac { y-2 }{ 5 } =\frac { z-3 }{ 4 } \)

Here \(P(A)=\frac { 1 }{ 7 } ,P(B)=\frac { 1 }{ 5 } \)
\(P(\overset { \_ }{ A) } =1-\frac { 1 }{ 7 } =\frac { 6 }{ 7 } \)
\(P(\overset { \_ }{ B } )=1-\frac { 1 }{ 5 } =\frac { 4 }{ 5 } \)
Required probability =\(P(A\overset { \_ }{ B } )+P(\overset { \_ }{ A } B)=P(A)P(\overset { \_ }{ B } )+P(\overset { \_ }{ A } )P(B)\)
= \(\frac { 1 }{ 7 } \times \frac { 4 }{ 5 } +\frac { 6 }{ 7 } \times \frac { 1 }{ 5 } \)
\(=\frac { 4 }{ 35 } +\frac { 6 }{ 35 } =\frac { 10 }{ 35 } =\frac { 2 }{ 7 } \)

Let 'p' be the probability of success and 'q' be tha probability of failure.
Then p + q = 1 and p = 2q
Solving, \(p=\frac { 2 }{ 3 } ,q=\frac { 1 }{ 3 } \) Also n = 6.
Required probability = \(P(X\ge 4)\)
= P(4)+P(5)+P(6)
= \(^6C_4\)q²p⁴+\(^6C_5\)q¹p⁵+\(^6C_6\)q⁰p⁶
= \(15{ \left( \frac { 1 }{ 3 } \right) }^{ 2 }{ \left( \frac { 2 }{ 3 } \right) }^{ 4 }+6{ \left( \frac { 1 }{ 3 } \right) }{ \left( \frac { 2 }{ 3 } \right) }^{ 5 }+\left( 1 \right) \left( 1 \right) { \left( \frac { 2 }{ 3 } \right) }^{ 6 }\)
= \({ \left( \frac { 2 }{ 3 } \right) }^{ 4 }\left[ \frac { 15 }{ 9 } +\frac { 4 }{ 3 } +\frac { 4 }{ 9 } \right] =\frac { 31 }{ 9 } { \left( \frac { 2 }{ 3 } \right) }^{ 4 }\)

Let 'x' and 'y' be the number of gold rings and chains respectively.
We have:
\(x\ge 0\)  ...(1) 
\(y\ge 0\)...(2)
\(x+y\le 24\)...(3) 
\(x+\frac { y }{ 2 } \le 16\) ...(4)
The objective function, or the profit, Z is:
Z = 300x + 190y ..(5)
We have to maximise Z subject to (1)-(4).
For solution set, we draw the lines:
x = 0, y = 0, x + y = 24, 2x + y = 32.
The lines x + y = 24 and 2x + y = 32 meet at E (8,16).

The shaded portion represents the feasible region, which is bounded.
Applying Corner Point Method, we have:

Hence, the maximum profit is Rs. 5,440 and it is obtained when 8 gold rings and 16 chains are manufactured.