Since the school is boys school, no student of the school can be sister of any student of the school. Hence, R = Ф, showing that R is the empty relation. It is also obvious that the difference between heights of any two students of the school has to be less than 3 meters. This shows that R '  = A × A is the universal relation.

Here, \(\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|\)

Minors of elements of first row are
\(M_{11}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1-0=1, \quad M_{12}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0-0=0\)
and \(M_{13}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=0-0=0\)
Minors of elements of second row are
\(\begin{aligned} & M_{21}=\left|\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right|=0-0=0 \\ & M_{22}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1-0=1 \text { and } M_{23}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|=0-0=0 \end{aligned}\)
Minors of elements of third row are
\(\begin{aligned} & M_{31}=\left|\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right|=0-0=0  M_{32}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|=0-0=0 \text { and } M_{33}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1-0=1 \end{aligned}\)
Hence, cofactors of elements of first row are
\(A_{11}=(-1)^{1+1} M_{11}=1 \times 1=1, A_{12}=(-1)^{1+2} M_{12}=-1 \times 0=0 \)
\(A_{13}=(-1)^{1+3} M_{13}=1 \times 0=0\)
cofactors of elements of second row are
\( A_{21}=(-1)^{2+1} M_{21}=1 \times 0=0, A_{22}=(-1)^{2+2} M_{22}=1 \times 1=1 \)
\(A_{23}=(-1)^{2+3} M_{23}=-1 \times 0=0 \)
Coafactors of elements of third row are
\( A_{31}=(-1)^{3+1} M_{31}=1 \times 0=0, A_{32}=(-1)^{3+2} M_{32}=-1 \times 0=0 \)
\(A_{33}=(-1)^{3+3} M_{33}=1 \times 1=1 \)

Here, \(\left|\begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array}\right|\)
Minors of elements of first row are
\(M_{11}=\left|\begin{array}{cc} 5 & -1 \\ 1 & 2 \end{array}\right|=10+1=11, M_{12}=\left|\begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array}\right|=6-0=6 \text { and } M_{13}=\left|\begin{array}{ll} 3 & 5 \\ 0 & 1 \end{array}\right|=3-0=3\)
Minors of elements of second row are
\(M_{21}=\left|\begin{array}{ll} 0 & 4 \\ 1 & 2 \end{array}\right|=0-4=-4 \)
\(M_{22}=\left|\begin{array}{ll} 1 & 4 \\ 0 & 2 \end{array}\right|=2-0=2 \text { and } M_{23}=\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|=1-0=1\)
Minors of elements of third row are
\(\begin{aligned} & M_{31}=\left|\begin{array}{cc} 0 & 4 \\ 5 & -1 \end{array}\right|=0-20=-20 \\ & M_{32}=\left|\begin{array}{cc} 1 & 4 \\ 3 & -1 \end{array}\right|=-1-12=-13 \text { and } M_{33}=\left|\begin{array}{ll} 1 & 0 \\ 3 & 5 \end{array}\right|=5-0=5 \end{aligned}\)
Hence, cofactors of elements of first row are
\(A_{11}=(-1)^{1+1} M_{11}=1 \times 11=11, A_{12}=(-1)^{1+2} M_{12}=-1 \times 6=-6 \)
\(A_{13}=(-1)^{1+3} M_{13}=1 \times 3=3\)
cofactors of elements of second row are
\(A_{21}=(-1)^{2+1} M_{21}=-1 \times-4=4, A_{22}=(-1)^{2+2} M_{22}=1 \times 2=2 \)
\(A_{23}=(-1)^{2+3} M_{23}=1 \times 1=-1\)
Coafactors of elements of third row are
\(A_{31}=(-1)^{3+1} M_{31}=1 \times-20=-20 \)
\( A_{32}=(-1)^{3+2} M_{32}=-1 \times-13=13 \)
\( A_{33}=(-1)^{3+3} M_{33}=1 \times 5=5\)

We have, \(2 X+3 Y=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]\) and
\(3 X+2 Y=\left[\begin{array}{rr} 2 & -2 \\ -1 & 5 \end{array}\right]\)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 , we get
\(4 X+6 Y=\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right]\) and \( 9 X+6 Y=\left[\begin{array}{rr}6 & -6 \\ -3 & 15\end{array}\right]\)
On subtracting Eq. (iii) from Eq. (iv), we get
\( (9 X+6 Y)-(4 X+6 Y)=\left[\begin{array}{rr} 6 & -6 \\ -3 & 15 \end{array}\right]-\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right] \)
\(\Rightarrow 9 X+6 Y-4 X-6 Y=\left[\begin{array}{rr} 6-4 & -6-6 \\ -3-8 & 15-0 \end{array}\right]=\left[\begin{array}{rr} 2 & -12 \\ -11 & 15 \end{array}\right] \)
\(\Rightarrow X=\frac{1}{5}\left[\begin{array}{rr}2 & -12 \\ -11 & 15\end{array}\right]=\left[\begin{array}{cc}\frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3\end{array}\right]\) 
On substituting the value of X in Eq. (i), we get
\( \Rightarrow\left[\begin{array}{cc} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{array}\right]+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \)
\(\Rightarrow \left[\begin{array}{cc} \frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6 \end{array}\right]+3 Y=\left[\begin{array}{cc} 2 & 3 \\ 4 & 0 \end{array}\right] \)
\(\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} 2-\frac{4}{5} & 3+\frac{24}{5} \\ 4+\frac{22}{5} & 0-6 \end{array}\right]\)
\(\Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{array}\right]=\left[\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{array}\right] \) 

\(\sin ^{-i}\left[\cos \left(\frac{33 \pi}{5}\right)\right]=\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right] \)
\(=\sin ^{-1}\left[\cos \frac{3 \pi}{5}\right] \quad[\because \cos (2 n \pi+\theta)=\cos \theta] \)
\(=\sin ^{-1}\left[\cos \left(\frac{\pi}{2}+\frac{\pi}{10}\right)\right]=\sin ^{-1}\left[-\sin \left(\frac{\pi}{10}\right)\right] \)
\(=-\sin ^{-1}\left(\sin \frac{\pi}{10}\right) \quad[1]\)
\(=-\frac{\pi}{10} \quad\left[\because \sin ^{-1}(\sin x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right][1] \)

|3A|=108
Alternative Method:
|kA|=kⁿ|A|
⇒ |3A|=3³(4)
=27(4)
=108

Any square matrix can be expressed
\(A=\frac{1}{2}\left(A+A^{\prime}\right)+\frac{1}{2}\left(A-A^{\prime}\right)\)
we know that (A + A′) is a symmetric matrix and (A – A′) is a skew symmetric matrix. Since for any matrix A, (kA)′ = kA′, it follows that \(\frac{1}{2}\left(A+A^{\prime}\right)\) is symmetric matrix and \(\frac{1}{2}\left(A-A^{\prime}\right)\) is skew symmetric matrix. Thus, any square matrix can be expressed as the sum of a symmetric and a skew symmetric matrix.

Let y = tan⁻¹(−1)
⇒ tan y = −1
We know that the range of the principal value branch of tan⁻¹x is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text {. }\)
\( \Rightarrow \tan y=\tan \left(\frac{-\pi}{4}\right) \)
\(\therefore y=\frac{-\pi}{4} \)
Hence, the principal value of tan⁻¹(−1) is \(\frac{-\pi}{4} \).

 The given system of equations can be written in the form of AX=B, where " id="MathJax-Element-1742-Frame" role="presentation" style="position: relative;" tabindex="0"> The given system of equations can be written in the form of AX=B, where 
\(A=\left[\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right], X=\left[\begin{array}{c} x \\ y \\ z \end{array}\right] \text { and } B=\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right]\)
\(\text { Now, }|A|=1(12-5)+1(9+10)+2(-3-8)=7+19-22=4 \neq 0\)
 Thus, A is non-singular. Therefore, its inverse exists. " id="MathJax-Element-1745-Frame" role="presentation" style="position: relative;" tabindex="0">Thus, A is non-singular. Therefore, its inverse exists. 
\(\text { Now, } A_{11}=7, A_{12}=-19, A_{13}=-11 \)
\(A_{21}=1, A_{22}=-1, A_{23}=-1 \)
\(A_{31}=-3, A_{32}=11, A_{33}=7\)
\(\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right] \)
\(\therefore X=A^{-1} B=\frac{1}{4}\left[\begin{array}{ccc} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{array}\right]\left[\begin{array}{c} 7 \\ -5 \\ 12 \end{array}\right] \)
\(\Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{4}\left[\begin{array}{c} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{array}\right] \)
\(=\frac{1}{4}\left[\begin{array}{l} 8 \\ 4 \\ 12 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right]\)
Hence, x = 2, y = 1and z = 3

Let \(x_{1}, x_{2} \in\left[0, \frac{\pi}{2}\right],\) such that \(x_{1} \neq x_{2}\)
Then, \(\sin x_{1} \neq \sin x_{2}\) and \(\cos x_{1} \neq \cos x_{2}\)
\(\cos x_{1} \neq \cos x_{2}\) for any two distinct values x in \(\cos x_{1} \neq \cos x_{2}\)
since function cannot give same value.
This is also true for cosine function] Hence, f and g must be one-one.
\(\left[\because a_{1} \neq a_{2} \Rightarrow f\left(a_{1}\right) \neq f\left(a_{2}\right)\right]\)
But \(\left.(f+g)^{*} 0\right)=\sin 0+\cos 0=1\)
and \((f+g)\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+\cos \frac{\pi}{2}=1\)
\(\therefore\)  f + g is not one-one.

Given
\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) =\frac { 1 }{ 2 } { tan }^{ -1 }x\)
\(\Rightarrow { tan }^{ -1 }1-{ tan }^{ -1 }x=\frac { 1 }{ 2 } { tan }^{ -1 }x\)
\(\Rightarrow \frac { 3 }{ 2 } { tan }^{ -1 }x=\frac { \pi }{ 4 } \)
\(\Rightarrow { tan }^{ -1 }x=\frac { \pi }{ 6 } \)
\(\Rightarrow x=tan\left( \frac { \pi }{ 6 } \right) \)
\(\therefore x=\frac { 1 }{ \sqrt { 3 } } \)

\(\mathrm{LHS}=(a I+b A)^{3}=\left\{\left[\begin{array}{ll} a & 0 \\ 0 & a \end{array}\right]+\left[\begin{array}{ll} 0 & b \\ 0 & 0 \end{array}\right]\right\}^{3} \)
\(=\left[\left.\begin{array}{ll} a & b \\ 0 & a \end{array}\right|^{3}=\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]\right. \)
\(=\left[\begin{array}{cc} a^{2} & 2 a b \\ 0 & a^{2} \end{array}\right]\left[\begin{array}{ll} a & b \\ 0 & a \end{array}\right]=\left[\begin{array}{cc} a^{3}+0 & a^{2} b+2 a^{2} b \\ 0+0 & 0+a^{3} \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{RHS}=a^{3} I+3 a^{2} b A=\left[\begin{array}{cc} a^{3} & 0 \\ 0 & a^{3} \end{array}\right]+\left[\begin{array}{cc} 0 & 3 a^{2} b \\ 0 & 0 \end{array}\right]=\left[\begin{array}{cc} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{array}\right] \)
\(\mathrm{LHS}=\mathrm{RHS} \)
\(\text {Hence, }(a I+b l)^{3}=a^{3} I+3 a^{2} b A \)

\( \tan ^{-1} \frac{x}{\sqrt{a^2-x^2}} \)
\( \text { Put } x=a \sin \theta \Rightarrow \frac{x}{a}=\sin \theta \Rightarrow \theta=\sin ^{-1}\left(\frac{x}{a}\right) \)
\( \therefore \tan ^{-1} \frac{x}{\sqrt{a^2-x^2}}=\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right) \)
\(=\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^2 \theta}}\right)=\tan ^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right. )\)
\( =\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1} \frac{x}{a} \)

(i) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right] \Rightarrow \mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right] \Rightarrow\left(\mathrm{A}^{\prime}\right)^{\prime}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right]=\mathrm{A}\)
Thus \(\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}\)
(ii) We have
\(\mathrm{A}=\left[\begin{array}{lll} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{array}\right], \mathrm{B}=\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right] \Rightarrow \mathrm{A}+\mathrm{B}=\left[\begin{array}{ccc} 5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{array}\right]\)
Therefore \((A+B)^{\prime}=\left[\begin{array}{cc} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)
Now \(\mathrm{A}^{\prime}=\left[\begin{array}{cc} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{array}\right], \mathrm{B}^{\prime}=\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right] \text {, }\)
So \(A^{\prime}+B^{\prime}=\left[\begin{array}{rr} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{array}\right]\)

Thus \((\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime} \)

(iii) We have
\(k B=k\left[\begin{array}{rrr} 2 & -1 & 2 \\ 1 & 2 & 4 \end{array}\right]=\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]\) 
and 
\((k B)^{\prime} =\left[\begin{array}{ccc} 2 k & -k & 2 k \\ k & 2 k & 4 k \end{array}\right]^{\prime}=\left[\begin{array}{cc} 2 k & k \\ -k & 2 k \\ 2 k & 4 k \end{array}\right].\)
\(=k\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{array}\right]=k B^{\prime} \) 
Thus, (kB)' = kB'

\( \operatorname{cosec}\left(\sin ^{-1} \frac{1}{5}\right)+\sec \left(\cos ^{-1} \frac{1}{3}\right)\)
\( =\operatorname{cosec}\left(\operatorname{cosec}^{-1} 5\right)+\sec \left(\sec ^{-1} 3\right)\)
\(\left[\because \sin ^{-1} \frac{1}{x}=\operatorname{cosec}^{-1} x \text { and } \cos ^{-1} \frac{1}{x}=\sec ^{-1} x\right] \)
\(=5+3=8 \quad\left[\because \operatorname{cosec}\left(\operatorname{cosec}^{-1} x\right)=x \text { and } \sec \left(\sec ^{-1} x\right)=x\right] \)

Set A is the set of all books in the library of a college.
R = {x, y): x and y have the same number of pages}
Now, R is reflexive since (x, x) ∈ R as x and x has the same number of pages.
Let (x, y) ∈ R ⇒ x and y have the same number of pages.
⇒ y and x have the same number of pages.
⇒ (y, x) ∈ R
∴ R is symmetric.
Now, let (x, y) ∈R and (y, z) ∈ R.
⇒ x and y and have the same number of pages and y and z have the same number of pages.
⇒ x and z have the same number of pages.
⇒ (x, z) ∈ R
∴ R is transitive.
Hence, R is an equivalence relation.

Let \(x_{ 1 },x_{ 2 }\in R\).
Now \(f(x_{ 1 })=f(x_{ 2 })\Rightarrow cosx_{ 1 }=cosx_{ 2 }\)
\(\Rightarrow \)  \(x_{ 1 }=(2n\pi +x_{ 2 })\)
\(\Rightarrow \) '\(f\)' is not one-one.
(ii) Since cos \(x\) lies in [-1,1],
\(\therefore \) R is not fully covered.
Hence, '\(f\)'  is not onto.

\(\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| \)
= \(\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right|\)
= 0

From the option, we can see only option (a) satisfy both the equations

Clearly,
\( f(a) =\left|\begin{array}{ccc} 0 & 0 & a-b \\ 2 a & 0 & a-c \\ a+b & a+c & 0 \end{array}\right| \)
\(=[(a-b)\{2 a \cdot(a+c)\}] \neq 0 \) 
\( \therefore \ f(b) =\left|\begin{array}{ccc} 0 & b-a & 0 \\ b+a & 0 & b-c \\ 2 b & b+c & 0 \end{array}\right|\)
\( =-(b-a)[2 b(b-c)] \)
\( =-2 b(b-a)(b-c) \neq 0 \) 
 

The number of elements in m x n matrix is equal to mn.

Cosine functions respected to any interval \(\left[-\pi, 0]\right.\)

Total number of functions = (n(B)^n(A) = 2ⁿ. Clearly a function will not be onto if all elements of A map to either a or b.

As -1 ≤ -x² < 1
⇒ 1 ≥ x² ≥ -1
⇒ 0 ≤ x² ≤ 1
⇒ |x| ≤ 1
⇒ -1 ≤ x ≤ 1.

Here (1,2) e R, (2,1) € R, if transitive (1,1) should belong to R.

We have, R = {(l₁,l₂) :l₁ is parallel to l₂}
(i) If l₁ is parallel to l₂, then l₂ is parallel to l_1.
So, if (l₁, l₂) \(\in R,\) then (l₂, l₁) \(\in R\)
\(\therefore\) R is symmetric.
(ii) If l₁ is parallel to l₂ and l₂ is parallel to l₃, then l₁ is parallel to l₃.
So, if \(\left(l_1, l_2\right) \in R,\left(l_2, l_3\right) \in R\), then (l₁,l₃)\(\in R\)
\(\therefore\) R is transitive.
(iii) Let equation of line parallel to y = 3x + 2 be y = mx + c, where m is the slope of line. Since, y = 3x + 2 and y = mx + c are parallel. Slope of (y =3x + 2) = Slope of(y = mx + c)
\(\Rightarrow\) 3 = m i.e. m = 3
Hence, the required line is
y = 3x + c, where c \(\in R\)
Or
We have, S = {(I₁, I₂) : l₁ is perpendicular to l₂}
For Symmetric If I₁, is perpendicular to I₂, then l₂ is perpendicular to l₁.
So, if (l₁,l₂) \(\in S\), then (l₂, l₁) \(\in S\)
\(\therefore\) S is symmetric.
For Transitive If I₁, is perpendicular to l₂, and l₂, is perpendicular to l₃, then I₁, is not perpendicular to l₃, it is parallel to I₃.
So, if \(\left(l_1, l_2\right) \in S,\left(l_2, l_3\right) \in S \text {, then }\left(l_1, l_3\right) \notin S \text {. }\)
\(\therefore\) S is not transitive.

(i) (a) : \(A B=\left[\begin{array}{cc} 2 & -1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 5 & 2 \\ 7 & 4 \end{array}\right]\) 
\(=\left[\begin{array}{cc} 10-7 & 4-4 \\ 15+28 & 6+16 \end{array}\right]=\left[\begin{array}{cc} 3 & 0 \\ 43 & 22 \end{array}\right]\) 
(ii) (c)
(iii) (b) : We have, CD - AB = 0
\(\Rightarrow\left[\begin{array}{ll} 2 & 5 \\ 3 & 8 \end{array}\right]\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]-\left[\begin{array}{cc} 3 & 0 \\ 43 & 22 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\) 
\(\Rightarrow\left[\begin{array}{ll} 2 a+5 c & 2 b+5 d \\ 3 a+8 c & 3 b+8 d \end{array}\right]-\left[\begin{array}{cc} 3 & 0 \\ 43 & 22 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\) 
\(\Rightarrow\left[\begin{array}{cc} 2 a+5 c-3 & 2 b+5 d \\ 3 a+8 c-43 & 3 b+8 d-22 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\) 
By equality of matrices, we get 2a + 5c - 3 = 0  ...(i)
3a + 8c - 43 = 0 ...(ii)
2b + 5d = 0 ...(iii)
3b + 8d - 22 = 0 ...(iv)
Solving (i) and (ii), we get a = -191, e = 77
(iv) (c) : Solving (iii) and (iv), we get b = -110, d = 44
(v) (d) : Wehave, \(B+D=\left[\begin{array}{ll} 5 & 2 \\ 7 & 4 \end{array}\right]+\left[\begin{array}{cc} -191 & -110 \\ 77 & 44 \end{array}\right]\) 
\(=\left[\begin{array}{cc} -186 & -108 \\ 84 & 48 \end{array}\right]\)