0 (Using properties)

(I + A)²-3A = I
Alternative Mwthod:
Given A² = A,
(I+A)²-3A = I² + 2IA + A²-3A
= I + 2A + A - 3A = I

We have, y = \(log\left( tanx\frac { x }{ 2 } \right) \)
\(\frac { dy }{ dx } =\frac { 1 }{ \frac { tanx }{ 2 } } \times { sec }^{ 2 }\frac { x }{ 2 } \times \frac { 1 }{ 2 } \)
\(\Rightarrow \frac { dy }{ dx } =\frac { \frac { cosx }{ 2 } }{ \frac { sinx }{ 2 } } \times \frac { 1 }{ { cos }^{ 2 }\frac { x }{ 2 } } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2sin\frac { x }{ 2 } cos\frac { x }{ 2 } } =\frac { 1 }{ sinx } \)
\(\Rightarrow \frac { dy }{ dx } =cosecx\)

We have, y = a^x +x^a+x^x+a^a
Let v = x^x
log v = x log x
\(\frac { 1 }{ v } \frac { dv }{ dx } =x+\frac { 1 }{ x } +logx\)
\(\frac { dv }{ dx } =v\left| 1+logx \right| \)
= x^x(1+lodx)
\(\frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+\frac { dv }{ dx } +0\)
\(\Rightarrow \frac { dy }{ dx } ={ a }^{ x }loga+a{ x }^{ a-1 }+{ x }^{ x }(1+logx)\)

\(Let\quad { sec }^{ -1 }x=\theta ,\quad then\quad x=sec\theta \quad and\quad for\quad x<-1,\)
\(\frac { \pi }{ 2 } <\theta <\pi \)
Given expression = cot^-1(-cot \(\theta \))
\(={ cot }^{ -1 }\left[ cot\left( \pi -\theta \right) \right] =\pi -{ sec }^{ -1 }x\quad as\quad 0<\pi -\theta <\frac { \pi }{ 2 } \)

Let A be a skew-symmetric matrix. Then by definition \({ A }^{ \prime }=-A\)
\(\Rightarrow\) the (i, j)^th element of \({ A }^{ \prime }\) = the (i, j)^th element of (- A)
\(\Rightarrow\) the (j, i)^th element of A = - the (i, j)^th element of A
For the diagonal elements i = j \(\Rightarrow\) the (i, j)the element of A = - the (i, j)^th element of A.
\(\Rightarrow\)  the (i, j)^th element of A = 0
Hence the diagonal elements are all zero.

(i) Let
A = (1,2,3)
A x A = (1, 1) (2, 2) (3, 3) (1, 2) (2, 1) (3, 2) (2, 3) (1, 3), (3, 1)
\(\therefore\) R = {(1, 1) (1, 2) (2, 1) (2, 2) (2, 3) (1,3) (3, 3) (3,2) (3, 1)}.
(ii) R = {(1, 1) (1,2) (2,1) (2,2) (2,3), (3, 3) (3,2)}

The given function is f: X → Y and relation on X is R={(a, b): f(a) = f (b)}
Reflexive Since, for every x ∈ X, we have
f'(x) = f(x)
⇒ (xx) ∈ R, ∀ ∈ X Therefore, R is reflexive.
Symmetric Let (x, y) ∈ R
Then, f(x)=f(y)
⇒ f(y)=f(x)
⇒ (y, x) ∈ R
Thus, (x, y) ∈ R ⇒  (y, x)∈ R, ∀x, y∈ X
Therefore, R is symmetric.
Transitive Let x, y, z∈ X such that
(x, y) ∈ R and (y, z) ∈ R
Given a relation 5 in \(N \times N\), defined as
(a, b) S(c, d), if a+d=b+c.
Reflexive Let (a, b) be any arbitrary element of \(N \times N\)
i.e. \((a, b) \in N \times N\), where \(a, b \in N\)
Now, as a+b=b+a
[∴ addition is commutative ]
Therefore \quad(a, b) S(a, b)
So, S is reflexive.
Symmetric \(\operatorname{Let}(a, b),(c, d) \in N \times N\), such that (a, b)
S(c, d). Then, a+d=b+c
\( \Rightarrow  b+c=a+d \Rightarrow c+b=d+a \)
\(\Rightarrow (c, d) S(a, b)\)
So, S is symmetric.
Transitive Let \((a, b),(c, d),(e, f) \in N \times N\) such that (a, b) S(c, d) and (c, d) S(e, f).
Then, a+d=b+c and c+f=d+e
On adding the above equations, we get
a+d+c+f=b+c+d+e 
\( \Rightarrow  a+f=b+e \Rightarrow(a, b) S(e, f)\)
So, S is transitive.
Thus, S is reflexive, symmetric and transitive. Hence, S is an equivalence

\({ tan }^{ -1 }\left[ \frac { cos\quad x }{ 1+sin\quad s } \right] \) \(\quad \because \) \(\begin{cases} cos\quad x={ cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } \\ and\quad 1+sin\quad x={ \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } \end{cases}\)
\(={ tan }^{ -1 }\left[ \frac { { cos }^{ 2 }\frac { x }{ 2 } -{ sin }^{ 2 }\frac { x }{ 2 } }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) \left( cos\frac { x }{ 2 } -sin\frac { x }{ 2 } \right) }{ { \left( cos\frac { x }{ 2 } +sin\frac { x }{ 2 } \right) }^{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ \frac { cos\frac { x }{ 2 } -sin\frac { x }{ 2 } }{ cos\frac { x }{ 2 } +sin\frac { x }{ 2 } } \right] Divide\quad by\quad cos\frac { x }{ 2 } ,\quad we\quad get\)
\(={ tan }^{ -1 }\left[ \frac { 1-tan\frac { x }{ 2 } }{ 1+tan\frac { x }{ 2 } } \right] \)
\(={ tan }^{ -1 }\left[ tan\left( \frac { \pi }{ 4 } -\frac { x }{ 2 } \right) \right] =\frac { \pi }{ 4 } -\frac { x }{ 2 } \)

We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

We have A + B - X = 0
By adding X on both the sides,
A + B - X + X = 0 + X
\(\Rightarrow\)  A + B = X
\(\Rightarrow X=\left[ \begin{matrix} 1 & 4 \\ 3 & 2 \\ 2 & 1 \end{matrix} \right] +\left[ \begin{matrix} 5 & 2 \\ -1 & 0 \\ 1 & 1 \end{matrix} \right] \)
\(\Rightarrow X=\left[ \begin{matrix} 6 & 6 \\ 2 & 2 \\ 3 & 2 \end{matrix} \right] \)

\(\frac{d y}{d x}=e^{x}+2 x e^{x^{2}}+3 x^{2} e^{x^{3}}+4 x^{3} e^{x^{4}}+5 x^{4} e^{x^{5}}\)

* is defined on Z⁺ as a * b = a - b
If a < b, then a−b ∉ Z⁺
a ∗ b ∉ Z⁺. Hence, * is not binary operation on Z⁺

We have:
\(f(x)=x^{ 2 }+2\) and \(g(x)=\frac { x }{ x-1 } \)
\(\therefore \) \(fog(x)=f(g(x))=(g(x))^{ 2 }+2\)
\(=\left( \frac { x }{ x-1 } \right) ^{ 2 }+2=\frac { x^{ 2 }+2(x-1)^{ 2 } }{ (x-1)^{ 2 } } \)
\(=\frac { x^{ 2 }+2(x^{ 2 }-2x+1) }{ (x-1)^{ 2 } } =\frac { 3x^{ 2 }-4x+2 }{ (x-1)^{ 2 } } \)
Hence, fog (2)\(=\frac { 3(4)-4(2)+2 }{ (2-1)^{ 2 } } =\frac { 12-8+2 }{ (1)^{ 2 } } =\frac { 6 }{ 1 } =6\)
And \(gof(x)=g(f(x))=\frac { f(x) }{ f(x)-1 } \)
\(=\frac { x^{ 2 }+2 }{ (x^{ 2 }+2)-1 } =\frac { x^{ 2 }+2 }{ x^{ 2 }+1 } \)
Hence, \(gof(-3)=\frac { (-3)^{ 2 }+2 }{ (-3)^{ 2 }+1 } =\frac { 9+2 }{ 9+1 } =\frac { 11 }{ 10 } \) .

We have: \(x={ ae }^{ \theta }\left( sin\theta -cos\theta \right) \)
\(y={ ae }^{ \theta }\left( sin\theta +cos\theta \right) \)
\(\frac { dx }{ d\theta } ={ ae }^{ \theta }\left( cos\theta +sin\theta \right) +{ ae }^{ \theta }\left( sin\theta -cos\theta \right) \)
\(=2{ ae }^{ \theta }sin\theta \)
\(\frac { dy }{ d\theta } ={ ae }^{ \theta }\left( cos\theta -sin\theta \right) +{ ae }^{ \theta }\left( sin\theta +cos\theta \right) \)
\(=2{ ae }^{ \theta }cos\theta \)
\(\frac { dy }{ dx } =\frac { dy/d\theta }{ dx/d\theta } =\frac { 2{ ae }^{ \theta }cos\theta }{ 2{ ae }^{ \theta }sin\theta } =\frac { cos\theta }{ sin\theta } =cot\theta \)
\(Hence,\quad \frac { dy }{ dx } =cot\frac { \pi }{ 4 } =1\)

We have: \(A=\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] .\)
\(\therefore \ A\prime =\left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right]\)
\(\therefore \ AA\prime =\left[ \begin{matrix} 3 & 1 & -1 \\ 0 & 1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 0 \\ 1 & 1 \\ -1 & 2 \end{matrix} \right] \)
\(=\begin{bmatrix} 9+1+1 & 0+1-2 \\ 0+1-2 & 0+1+4 \end{bmatrix}=\begin{bmatrix} 11 & -1 \\ -1 & 5 \end{bmatrix}.\)
Which is a symmetric matrix.

Since A and B are both symmetric matrices, therefore A′ = A and B′ = B.
Let AB be symmetric, then (AB)′ = AB
But (AB)′ = B′A′= BA 
Therefore BA = AB
Conversely, if AB = BA, then we shall show that AB is symmetric.
Now (AB)′ = B′A′
= BA (as A and B are symmetric)
= AB
Hence AB is symmetric.

\(A=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right] \)
\(A^{2}=A \cdot A=\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]=\left[\begin{array}{cc} 9-1 & 3+2 \\ -3-2 & -1+4 \end{array}\right]=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right] \)
\(\therefore A^{2}-5 A+7 I \)
\(=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-5\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]+7\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \)
\(=\left[\begin{array}{rr} 8 & 5 \\ -5 & 3 \end{array}\right]-\left[\begin{array}{ll} 15 & 5 \\ -5 & 10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] \)
\(=\left[\begin{array}{ll} -7 & 0 \\ 0 & -7 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \)
\(\text { Hence, } A^{2}-5 A+7 I=O .\)
\(\therefore A \cdot A-5 A=-7 I \)
\(\Rightarrow A \cdot A\left(A^{-1}\right)-5 A A^{-1}=-7 I A^{-1} \quad\left[\text { Post-multiplying by } A^{-1} \text { as }|A| \neq 0\right] \)
\(\Rightarrow A\left(A A^{-1}\right)-5 I=-7 A^{-1} \)
\(\Rightarrow A I-5 I=-7 A^{-1} \)
\(\Rightarrow A^{-1}=-\frac{1}{7}(A-5 I) \)
\(\Rightarrow A^{-1}=\frac{1}{7}(5 I-A) \)
\(=\frac{1}{7}\left(\left[\begin{array}{ll} 5 & 0 \\ 0 & 5 \end{array}\right]-\left[\begin{array}{rr} 3 & 1 \\ -1 & 2 \end{array}\right]\right)=\frac{1}{7}\left[\begin{array}{lr} 2 & -1 \\ 1 & 3 \end{array}\right] \)
\(\therefore A^{-1}=\frac{1}{7}\left[\begin{array}{rr} 2 & -1 \\ 1 & 3 \end{array}\right]\)

(i) \(M_{11}=3, M_{12}=0, M_{21}=-4, M_{22}=2, \)
\(C_{11}=3, C_{12}=0, C_{21}=4 \text { and } C_{22}=2 \)
(ii) \(M_{11}=d, M_{12}=b, M_{21}=c \text { and } M_{22}=a \)
\(C_{11}=d, C_{12}=-b, C_{21}=-c, C_{22}=a\)

Take x = cos θ, then proceeding as above, we get, \({ \sin }^{ -1 }(2x\sqrt { 1-{ x }^{ 2 } } )\) = 2 cos^–1 x

Let \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) =y\), Then \(\sin y=\frac{1}{\sqrt{2}}\)
We know that the range of the principal value branch of \(\sin ^{-1} \text { is }\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) and \(\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\)
Therefore, principal value of \({ sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 2 } } \right) is \ \frac { \pi }{ 4 } \)

\({ A }^{ -1 }=\frac { 1 }{ 8 } \left[ \begin{matrix} 5 & -1 \\ -7 & 3 \end{matrix} \right] \)

\(\Delta =\left| \begin{matrix} \alpha & \beta & \gamma \\ { \alpha }^{ 2 } & { \beta }^{ 2 } & { \gamma }^{ 2 } \\ \beta +\gamma & \gamma +\alpha & \alpha +\beta \end{matrix} \right| \)
Using R1⟶R1+R3
=  \(\left| \begin{matrix} \alpha +\beta +\gamma & \gamma +\alpha +\beta & \alpha +\beta +\gamma \\ { \alpha }^{ 2 } & { \beta }^{ 2 } & { \gamma }^{ 2 } \\ \beta +\gamma & \alpha -\beta & \alpha -\gamma \end{matrix} \right| \)
Taking \(\alpha +\beta +\gamma \) common from R1
=  \(\alpha +\beta +\gamma )\left| \begin{matrix} 1 & 0 & 0 \\ { \alpha }^{ 2 } & { \beta }^{ 2 }-{ \alpha }^{ 2 } & { \gamma }^{ 2 }-{ \alpha }^{ 2 } \\ \beta +\gamma & \alpha -\beta & \alpha -\gamma \end{matrix} \right| \)
Using C2⟶C2-C1, C3⟶C3-C1
= \((\alpha +\beta +\gamma )(\beta -\alpha )(\gamma -\alpha )\left| \begin{matrix} 1 & 0 & 0 \\ { \alpha }^{ 2 } & \beta +\alpha & \gamma +\alpha \\ \beta +\gamma & -1 & -1 \end{matrix} \right| \)
Expanding with respect to R1
\(\Delta =(\alpha +\beta +\gamma )(\beta -\alpha )(\gamma -\alpha )(-\beta -\alpha +\gamma +\alpha )\)
\(\Delta =(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )(\alpha +\beta +\gamma )\)

\(tan^{ -1 }\left( \frac { x+1 }{ x-1 } \right) +tan^{ -1 }\left( \frac { x-1 }{ x } \right) =-tan^{ -1 }7\)
\(\Rightarrow tan^{ -1 }\left( \frac { \left( \frac { x+1 }{ x-1 } \right) +\left( \frac { x-1 }{ x } \right) }{ 1-\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) } \right) =-tan^{ -1 }7\)
if \(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) <1\)
\(\Rightarrow tan^{ -1 }\left[ \frac { x(x+1)+(x-1)^{ 2 } }{ \left( x-1 \right) x-\left( x+1 \right) \left( x-1 \right) } \right] =tan^{ -1 }7\)
\(\Rightarrow \frac { \left( x^{ 2 }+x \right) +\left( x^{ 2 }+1-2x \right) }{ \left( x^{ 2 }-x \right) -\left( x^{ 2 }-1 \right) } =tan\left[ -tan^{ -1 }7 \right] \)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \frac { 2x^{ 2 }-x+1 }{ -x+1 } =-7\)
\(\Rightarrow \) 2x² - 8x + 8 = 0
\(\Rightarrow \) (x - 2)² = 0
\(\Rightarrow \) x = 2
Let us now verify whether x = 2 satisfies the condition (i)
\(\left( \frac { x+1 }{ x-1 } \right) \left( \frac { x-1 }{ x } \right) =3\times \frac { 1 }{ 2 } =\frac { 3 }{ 2 } \) Which is not less than 1.
Hence this value does not satisfy the condition (i) there is no solution to the given trigonometric equation.

LHS = \(\frac { 1 }{ abc } \left| \begin{matrix} { (a+b) }^{ 2 } & c & c \\ a & { (b+c) }^{ 2 } & a \\ b & b & { (c+a) }^{ 2 } \end{matrix} \right| \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\)
\(=\frac { 1 }{ abc } \left| \begin{matrix} (a+b+c)(a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c+a)(b+c-a) & { a }^{ 2 } \\ (b+c+a)(b-c-a) & (b+c+a)(b-c-a) & { (c+a) }^{ 2 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }-{ R }_{ 2 }\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (ac+bc-{ c }^{ 2 }) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\(({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 3 })\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abcca } \left| \begin{matrix} (ac+bc) & { c }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & (ba+ca) & { a }^{ 2 } \\ 0 & 0 & { 2ca } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2{ c }^{ 2 }{ a }^{ 2 } }{ abcca } \left| \begin{matrix} (a+b) & c & c \\ a & (b+c) & a \\ 0 & 0 & { 1 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2 }{ b } (ab+ac+{ b }^{ 2 }+bc-ac)\)
\(=2{ (a+b+c) }^{ 3 }\)

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.

y = tan^-1 \(\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) \) 
= tan^-1 \(\left( \frac { \pi }{ 4 } \right) -{ tan }^{ -1 }{ x }^{ 2 }\)
\({ y }^{ ' }=0-\frac { -1 }{ 1+{ x }^{ 4 } } 2x=\frac { -2x }{ 1+{ x }^{ 4 } } \)