Let A and B be the points (2, 3, 1) and (3,-1, 2).
The direction cosines of\(\overrightarrow{AB},\) are \({1\over \sqrt{18}},{-4\over \sqrt{18}},{1\over \sqrt{18}}\).
But any point can be taken as first point.
Hence we have another set of direction cosines with opposite direction.
Thus, we have another set of direction ratio \({-1\over \sqrt{18}},{4\over \sqrt{18}},{-1\over \sqrt{18}}.\)

\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}-4\hat{j}+\hat{k}\)
Direction cosines are \({1\over \sqrt{18}},{-4\over \sqrt{18}},{1\over \sqrt{18}}\).

The condition is cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) = 1
Here \(\alpha =30^o,\beta =45^o,\gamma=60^o\)
cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) \(={3\over 4}+{1\over2}+{1\over4}\neq1.\)
There fore they are not direction angles of any vector.

If we denote the given expression by A, then using the scalar multiplication rule, we get
A =\( \begin{bmatrix} sec^2 \theta & sec\theta tan\theta \\ sec\theta tan\theta & sec^2\theta \end{bmatrix}\)-\( \begin{bmatrix} tan^2 \theta & tan\theta sec\theta \\ sec\theta tan\theta & tan^2\theta \end{bmatrix}\)=\(\begin{bmatrix}1&0 \\ 0 & 1 \end{bmatrix}\)

(i) Let A be the event ofgetting one mango and one apple, Then
\(P(A)=\frac{5 C_1 \times 4 C_1}{9 C_2}=\frac{5}{9}\)

(ii) Let B and C be the events of getting both are mango and both are an apple respectively then
\(P(\text { Bor } C) =P(B)+P(C) \)
\(=\frac{5 C_2}{9 C_2}+\frac{4 C_2}{9 C_2}=\frac{5 \times 4}{9 \times 8}+\frac{4 \times 3}{9 \times 8} \)
\(=\frac{8}{18}=\frac{4}{9}\)

Let   \(2\hat{i}+3\hat{j}-6\hat{k}\) and \(6\hat{i}-3\hat{j}+2\hat{k}\)
Let \(\theta \) be the angle between the given vectors.
\(\overrightarrow{a}.\overrightarrow{b}=(2\hat{i}+3\hat{j}-6\hat{k}).(6\hat{i}-3\hat{j}+2\hat{k})\)
= 12 - 9 - 12 = -9
\(|\overrightarrow{a}|=\sqrt{2^2+3^2+(-6)^2}=\sqrt{4+9+36}=\sqrt{49}=7\)
and \(|\overrightarrow{b}|=\sqrt{6^2+(-3)^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7\)
\(\therefore cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{a}|.|\overrightarrow{b}|}={-9\over 7(7)}={-9\over 49}\)
\(\Rightarrow \theta =cos^{-1}({-9\over 49})\)

Given  | \(\overrightarrow{a}\) | = 10, | \(\overrightarrow{b}\) | = 15 and \(\overrightarrow{a}\).\(\overrightarrow{b}\) = 75 \(\sqrt{2}\)
Let \(\theta\) be the angle between the vector \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

\(\theta ={\pi\over 4}.\)

Given a_ij = i-j
Let A = [a_ij] _{3 \(\times\) 3}
Ingeneralwe can wrlte \(\Lambda=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{21} & a_{13} \\ a_{31} & a_{32} & a_{33} \end{array}\right]\)
\(a_{i j}=1-j\)
\(\therefore\) a₁₁ = 1 - 1 = 0 a₂₁ = 2 - 1 = 1  a₃₁ = 3 - 1 = 2
a₁₂ = 1 - 2 = 0  a₂₂ = 2 - 2 = 0 a₃₂ = 3 - 2 = 1
a₁₃ = 1 - 3 = -2 a₂₃ = 2 - 3 = -1 a₃₃ = 3 - 3 = 0
\(\Rightarrow A=\left[ \begin{matrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow A^{T}=\left[ \begin{matrix} 0 & -1 & 2 \\ -1 & 0 & 1 \\ -2 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -1 & -2 \\ 1 & 0 & -1 \\ 2 & 1 & 0 \end{matrix} \right] =-A\)
Since A^T = -A  \(\Rightarrow\) A is a skew-symmetric matrix.

Let B = \(\left[ \begin{matrix} 0 & p & 3 \\ 2 & q^{ 2 } & -1 \\ r & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow { B }^{ T }=\left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] \)
Since B is a skew-symmetric matrix,
B^T = -B
\(\left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & p & 2 \\ 2 & { q }^{ 2 } & -1 \\ r & 1 & 0 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 0 & 2 & r \\ p & { q }^{ 2 } & 1 \\ 3 & -1 & 0 \end{matrix} \right] =-\left[ \begin{matrix} 0 & -p & -3 \\ -2 & -{ q }^{ 2 } & -1 \\ -r & -1 & 0 \end{matrix} \right] \)
Equating the corresponding entries on both sides, we get
-p = 2
p = -2
r = -3
q²  = -q²
q²  + q² = 0
 2q² = 0
 q² = 0
 q = 0
\(\therefore\) p = -2
r = -3
q = 0

\(A=\left[\begin{array}{cc} 4 & 2 \\ -1 & x \end{array}\right] ; I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
\(A-2 I=\left[\begin{array}{cc} 4 & 2 \\ -1 & x \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4-2 & 2+0 \\ -1+0 & x-2 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2 & 2 \\ -1 & x-2 \end{array}\right]\)
\(A-3 I=\left[\begin{array}{rr} 4 & 2 \\ -1 & x \end{array}\right]-3\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{cc} 4-3 & 2+0 \\ -1+0 & x-3 \end{array}\right]\)
\(=\left[\begin{array}{cc} 1 & 2 \\ -1 & x-3 \end{array}\right]\)
Given, \((A-2 I)(A-3 I)=\left[\begin{array}{cc} 2 & 2 \\ -1 & x-2 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ -1 & x-3 \end{array}\right]\)
\(=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(\left[\begin{array}{cc} 2-2 & 4+2 x-6 \\ -1-x+2 & -2+(x-2)(x-3) \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\)
\(-1-x+2=0\)
\(-x+1=0\)
\(x=1\)

Given A_{\(\alpha \)} = \(\begin{bmatrix} cos \alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
\({ A }_{ \alpha }^{ T }=\left[ \begin{matrix} cos\alpha & -sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right] \)
Also, it is given that \({ A }_{ \alpha }+{ A }_{ \alpha }^{ T }=1\)
\(\Rightarrow \left[ \begin{matrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{matrix} \right] +\left[ \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 2cos\alpha & 0 \\ 0 & 2cos\alpha \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
Equating the corresponding entries on both sides, we get
2 cos\(\alpha\) = 1 \(\Rightarrow\) \(cos\alpha =\frac { 1 }{ 2 } =cos\frac { \pi }{ 3 } \)
\(\alpha =2n\pi \pm \frac { \pi }{ 3 } ,n\epsilon z\)  \(\begin{aligned} {[\because \cos \theta} &=\cos \alpha \ \theta =2 n \pi \pm \alpha \forall n \in z] \end{aligned}\)
\(\alpha=2 n \pi \pm \frac{\pi}{3} \forall n \in z\)

A -B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 5 & 3 \\ -3 & 2 & 1 \\1 & 1 & 1 \end{bmatrix}\)
(A - B)^T =\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\) ..(1)
A^T - B^T =\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)-\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & -3 & 1 \\ 5 & 2 & 1 \\3 & 1 & 1 \end{bmatrix}\).....(2)
From (1) and (2), (A - B)^T= A^T - B^T.

A + B =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)+\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 7 & 1 \\ 3 & 0 & 9 \\ -1 & 5 & 3 \end{bmatrix}\)
(A+B)^T =\(\begin{bmatrix} 4 & 3 & -1 \\ 7 & 0 & 5 \\ 1 & 9 & 3 \end{bmatrix}\) .....(1)
A^T + B^T= \(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)+\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 4 & 3 & -1 \\ 7 & 0 & 5 \\ 1 & 9 & 3 \end{bmatrix}\).......(2)
From (1) and (2), (A+B)^T = A^T+ B^T.

AB =\(\begin{bmatrix} 4 & 6 & 2 \\ 0 & 1 & 5 \\ 0 & 3 & 2 \end{bmatrix}\)\(\begin{bmatrix} 0 & 1 & -1 \\ 3 & -1 & 4 \\ -1 & 2 & 1 \end{bmatrix}\)=\(\begin{bmatrix} 16 & 2 & -1 \\ -2 & 9 & 9 \\ 7 & 1 & 14 \end{bmatrix}\)
(AB)^T =\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(1)
B^T = \(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\), A^T=\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)
B^TA^T=\(\begin{bmatrix} 0 & 3 & -1 \\ 1 & -1 & 2 \\ -1 & 4 & 1 \end{bmatrix}\)\(\begin{bmatrix} 4 & 0 & 0 \\ 6 & 1 & 3 \\ 2 & 5 & 2 \end{bmatrix}\)=\(\begin{bmatrix} 16 & -2 & 7 \\ 2 & 9 & 1 \\ 22 & 9 & 14 \end{bmatrix}\)...(2)
From (1) and (2), (AB)^T = B^TA^T.

A² = AA =\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)=\(\begin{bmatrix} C_{11} &C_{12} &C_{13} \\ C_{21} & C_{22} &C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}\)

Consider the product \(\begin{vmatrix} a_1 &b_1 &c_1 \\ a_2 & b_2 &c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}\)\(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\)
= \(\begin{vmatrix} a_1A_1+b_1B_1+c_1C_1 & a_1A_2+b_1B_2+c_1C_2& a_1A_3+b_1B_3+c_1C_3\\ a_2A_1+b_2B_1+c_2C_1 & a_2A_2+b_2B_2+c_2C_2 &a_2A_3+b_2B_3+c_2C_3 \\ a_3 A_1+b_3B_1+c_3C_1 & a_3A_2+b_3B_2+c_3C_2 &a_3A_3+b_3B_3+ c_3C_3 \end{vmatrix}\)
= \(\begin{vmatrix} |A| &0 &0 \\ 0 & |A| &0 \\ 0 & 0 & |A| \end{vmatrix}\) = |A|³.
That is, |A| \(\times\) \(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\) = |A|³
\(\Rightarrow\)\(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\) = |A|².

Given A and B be two symmetric matrices.
\(\Rightarrow A^T=A \text { and } B^T=B\)
Given AB = BA.
\(\Leftrightarrow(A B)^T=B^T A^T\)
\(\Leftrightarrow(A B)^T=B A\)
\(\Leftrightarrow(A B)^T=A B\)
\(\Leftrightarrow A B \text { is a symmetric matrix. }\)

Let A =\(\begin{bmatrix} 1&2 &3 \\ 4 & 5 &6 \\ 7 & 8 & 9 \end{bmatrix}\)
= 1(45 - 48) -2(36 - 42) + 3(32 - 35)
= I(-3) - 2(-6) + 3(-3)
= -3 + 12 - 9
\(|A|=0\)
\(\therefore\) A is singular

\((\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b}) =3(\vec{a} \times \vec{a})-\vec{a} \times \vec{b}+9 \vec{b} \times \vec{a}-3 \vec{b} \times \vec{b} \)
\(=\overrightarrow{0}+\vec{b} \times \vec{a}+9(\vec{b} \times \vec{a})-\overrightarrow{0} \)
\(=10 \vec{b} \times \vec{a} \)
\(\frac{\sqrt{3}}{2} \times 2 =|\vec{a} \times \vec{b}| \)
\(|\vec{a} \times \vec{b}| =\sqrt{3} \)
\([(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})]^{2} =100(\vec{b} \times \vec{a})^{2} \)
\(=100(\vec{a} \times \vec{b})^{2}=100(\sqrt{3})^{2} \)
\(=100 \times 3=300 \)

\(\text { W.K.T }|\vec{a} \cdot \vec{b}|^{2}+|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\)
\(|\vec{a} \times \vec{b}|^{2} =13^{2} \cdot 5^{2}-|\vec{a} \cdot \vec{b}|^{2} \)
\(=(169)(25)-3600=4225-3600 =625 \)
\(|\vec{a} \times \vec{b}| =25 \)

\(\underbrace{\overrightarrow{A B}}+ \overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A C}+\overrightarrow{C D}}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A D}+\overrightarrow{D A}} \)
\(=\overrightarrow{A A}=\overrightarrow{0} . \)

\(\left|\begin{array}{ccc} a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b \end{array}\right| =0 \)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{lll} a & 2 b & 2 c \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{2} \rightarrow 2 R_{2} \)
\(\frac{1}{2}\left|\begin{array}{ccc} a-6 & 0 & 0 \\ 6 & 2 b & 2 c \\ 4 & a & b \end{array}\right| =0 \quad R_{1} \rightarrow R_{1}-R_{2} \)
\(\Rightarrow \frac{1}{2}\left[(a-6)\left(2 b^{2}-2 a c\right)\right] =0 \)
\((a-6)\left(2 b^{2}-2 a c\right) =0 \)
\(a=6,2 b^{2} =2 a c \)
\(b^{2} =a c \quad \therefore b^{3}=a b c \)

\(|A|=\left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & c \\ b & -c & 0 \end{array}\right|\)
\(=0-a(-b c)-b(a c)=a b c-a b c=0\)

\(2 X=\left[\begin{array}{ll} 3 & 8 \\ 7 & 2 \end{array}\right]-\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & 6 \\ 4 & -2 \end{array}\right]\)
\(X=\left[\begin{array}{cc} 1 & 3 \\ 2 & -1 \end{array}\right]\)

\(A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \)
\(a_{11}=\frac{1}{2}(3-2)=\frac{1}{2} ; a_{12}=\frac{1}{2}(3-4)=\frac{-1}{2} \)
\(a_{21}=\frac{1}{2}(3(2)-2)=\frac{4}{2}=2 ; a_{22}=\frac{1}{2}(6-4)=\frac{2}{2}=1 \)
\(\therefore A=\left[\begin{array}{ll} \frac{1}{2} & -\frac{1}{2} \\ 2 & 1 \end{array}\right] \)