(i) (b) : If Rs. 15000 is invested in bond X, then the amount invested in bond Y = Rs. (35000 - 15000) = Rs. 20000.

and

(ii) (c) : The amount of interest received on each bond is given by
\(A B=\left[\begin{array}{ll} 15000 & 20000 \end{array}\right] \times\left[\begin{array}{c} 0.1 \\ 0.08 \end{array}\right]\) 
= [15000 x 0.1 + 20000 x 0.08] = [1500 + 1600] = 3100
(iii) (c) : Let Rs. x be invested in bond X and then Rs. (35000 - x) will be invested in bond Y.
Now, total amount of interest is given by 
\(\left[\begin{array}{ll} x & 35000-x \end{array}\right]\left[\begin{array}{c} 0.1 \\ 0.08 \end{array}\right]=[0.1 x+(35000-x) 0.08]\) 
But, it is given that total amount of interest = Rs. 3200
\(\therefore\) 0.1x + 2800 - 0.08x = 3200
\(\Rightarrow 0.02 x=400 \Rightarrow x=20000\) 
Thus, Rs. 20000 invested in bond X and Rs. 35000 - Rs. 20000
= Rs. 15000 invested in bond Y.
(iv) (a) : AB will give the total amount of interest received on both bonds.
(v) (b) : Let Rs x invested in bond X, then we have
\(x \times \frac{10}{100}=500 \Rightarrow x=5000\) 
Thus, amount invested in bond X is Rs.5000 and so investment in bond Y be Rs. (35000 - 5000) = Rs. 30000

(i) (a) : Number of items purchased by shopkeepers A, B and C can be written in matrix form as

(ii) (b) : Since \(Y=\left[\begin{array}{c} 40 \\ 12 \\ 3 \end{array}\right]\)\(Note book\\ Pen\\ Penc\) 
\(\therefore \quad X Y=\left[\begin{array}{lll} 144 & 60 & 72 \\ 120 & 72 & 84 \\ 132 & 156 & 96 \end{array}\right]\left[\begin{array}{c} 40 \\ 12 \\ 3 \end{array}\right]\) 
\(=\left[\begin{array}{c} 5760+720+216 \\ 4800+864+252 \\ 5280+1872+288 \end{array}\right]=\left[\begin{array}{l} 6696 \\ 5916 \\ 7440 \end{array}\right]\) 
(iii) (d) : Bill of A is Rs.6696.
(iv) (c) : (A + I)² = A² + 2A + 1= 3A + I
\(\Rightarrow\) (A + I)³ = (3A + I) (A + I)
= 3A² + 4A + I = 7A + I
\(\therefore\) (A+I)³-7A = I
(v) (d) : A ² - B² = (A - B) (A + B) = A² + AB - BA - B²
\(\therefore\) AB=BA. 

(i) (a) : Let F be the matrix representing the number of family members and R be the matrix representing the requirement of calories and proteins for each person. Then
 
 
(ii) (b) : The requirement of calories and proteins for each of the two- families is given by the product matrix FR.
\(F R=\left[\begin{array}{lll} 4 & 4 & 4 \\ 2 & 2 & 2 \end{array}\right]\left[\begin{array}{ll} 2400 & 45 \\ 1900 & 55 \\ 1800 & 33 \end{array}\right]\) 
\(=\left[\begin{array}{ll} 4(2400+1900+1800) & 4(45+55+33) \\ 2(2400+1900+1800) & 2(45+55+33) \end{array}\right]\) 
 
(iii) (c)
(iv) (c) : Since, AB = B ...(i) and BA = A ...(ii)
\(\therefore\) A² + B² = A·A + B·B 
= A(BA) + B(AB) [using (i) and (ii)]
= (AB)A + (BA)B [Associative law]
= BA +AB [using (i) and (ii]
= A+B

(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, \(B+D=\left[\begin{array}{ll} 5 & 2 \\ 7 & 4 \end{array}\right]+\left[\begin{array}{cc} -191 & -110 \\ 77 & 44 \end{array}\right]\) 
\(=\left[\begin{array}{cc} -186 & -108 \\ 84 & 48 \end{array}\right]\)

Let A be the 2 x 3 matrix representing the annual sales of products in two markets.

Let B be the column matrix representing the sale price of each unit of products x, y, z.
\(\therefore \quad B=\left[\begin{array}{c} 2.5 \\ 1.5 \\ 1 \end{array}\right]\) 
Now, revenue = sale price x number of items sold
\(=\left[\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}\right]\left[\begin{array}{c} 2.5 \\ 1.5 \\ 1 \end{array}\right]\) 
\(=\left[\begin{array}{l} 25000+3000+18000 \\ 15000+30000+8000 \end{array}\right]=\left[\begin{array}{l} 46000 \\ 53000 \end{array}\right]\) 
Therefore, the revenue collected from Market I = Rs.46000 and the revenue collected from Market II = Rs. 53000.
(i) (c)
(ii) (b)
(iii) (d) : Let C be the column matrix representing cost price of each unit of products x, y, z.
Then, \(C=\left[\begin{array}{c} 2 \\ 1 \\ 0.5 \end{array}\right]\) 
\(\therefore\) Total cost in each market is given by 
\(A C=\left[\begin{array}{ccc} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{array}\right]\left[\begin{array}{c} 2 \\ 1 \\ 0.5 \end{array}\right]\) 
\(=\left[\begin{array}{c} 20000+2000+9000 \\ 12000+20000+4000 \end{array}\right]=\left[\begin{array}{l} 31000 \\ 36000 \end{array}\right]\) 
Now, Profit matrix = Revenue matrix, Cost matrix
\(=\left[\begin{array}{l} 46000 \\ 53000 \end{array}\right]-\left[\begin{array}{l} 31000 \\ 36000 \end{array}\right]=\left[\begin{array}{l} 15000 \\ 17000 \end{array}\right]\) 
Therefore, the gross profit from both the markets
= Rs.15000+ Rs.17000= Rs.32000
(iv) (a) : We have \(A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\) 
\(\therefore \quad A^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I\) 
(v) (a) : We have, (AB' -BA')' = (B')'A' - (A')'B'
= BA' - AB' = -(AB' - BA')
Thus, AB' - BA' is a skew-symmetric matrix.

Combined sales in September and October for each farmer in each variety is given by

(i) (c) : Combined sales of Masoor in September and October for farmer Balwan Singh = Rs. 40000
(ii) (d) : Combined sales of Urad in September and October for farmer Shyam = Rs. 15000
(iii) (a) : Change in sales from September to October is given by
 
\(\therefore\) Decrease in sales of Mung from September to October for farmer Shyam = Rs. 24000.
(iv) (b) : Required profit is given by
\(2 \% \text { of } B=\frac{2}{100} \times B=0.02 \times B\) 

 
thus,in October Shyam receives Rs. 100, Rs. 200 and Rs. 120 as profit in the sale of each variety of pulses, respectively and Balwan Singh receives a profit of Rs. 400, Rs. 200 and Rs. 200 in the sale of each variety of pulses respectively. 

(i) (a) : Clearly,

(ii) (d) : Since Q is a 3 x 1 matrix, therefore
 
(iii) (a) : Clearly, total funds collected by each school is given by the matrix.
\(P Q=\left[\begin{array}{ccc} 40 & 50 & 20 \\ 25 & 40 & 30 \\ 35 & 50 & 40 \end{array}\right]\left[\begin{array}{c} 25 \\ 100 \\ 50 \end{array}\right]\) 
\(=\left[\begin{array}{c} 1000+5000+1000 \\ 625+4000+1500 \\ 875+5000+2000 \end{array}\right]=\left[\begin{array}{l} 7000 \\ 6125 \\ 7875 \end{array}\right]\) 
\(\therefore\) Funds collected by school A is Rs. 7000
Funds collected by school B is Rs. 6125
Pimds collected by school C is Rs. 7875
(iv) (b) 
(v) (b) : Total funds collected for the required purpose
= Rs. (7000 + 6125 + 7875) = Rs. 21000

(i) (b) : In 2019,dealer A sold 120 Hatchback, 50 Sedan and 10 SUV; dealer B sold 100 Hatchback, 30 Sedan and 5 SUV and dealer C sold 90 Hatchback, 40 Sedan and 2 SUV
\(\therefore\)  Required matrix, say P, is given by
 
(ii) (a) : In 2020,
dealer A sold 300 Hatchback, 150 Sedan, 20 SUV
dealer B sold 200 Hatchback, 50 sedan, 6 SUV
dealer C sold 100 Hatchback, 60 sedan,S SUV
\(\therefore\) Required matrix, say Q, is given by
 
(iii) (c) : Total number of cars sold in two given years, by each dealer, is given by
 
 
(iv) (c): The increase in sales from 2019 to 2020 is given by
 
 
(v) (c) : The amount of profit in 2020 received by each dealer is given by the matrix
 
\(\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{c} 15000000+15000000+4000000 \\ 10000000+5000000+1200000 \\ 5000000+6000000+1000000 \end{array}\right]\) 
\(\begin{array}{r} A \\ =B \\ C \end{array}\left[\begin{array}{l} 34000000 \\ 16200000 \\ 12000000 \end{array}\right]\)

(i) (c) : Let Rs. A, Rs. B and Rs.C be the cost incurred by the organisation for villages X, Y and Z respectively. Then A, B, C will be given by the following matrix equation.
\(\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\left[\begin{array}{l} 50 \\ 20 \\ 40 \end{array}\right]=\left[\begin{array}{c} A \\ B \\ C \end{array}\right]\) 
\(\Rightarrow\left[\begin{array}{l} A \\ B \\ C \end{array}\right]=\left[\begin{array}{c} 400 \times 50+300 \times 20+100 \times 40 \\ 300 \times 50+250 \times 20+75 \times 40 \\ 500 \times 50+400 \times 20+150 \times 40 \end{array}\right]\) 
(ii) (c)
(iii) (b)
(iv) (c) : Total number of toilets that can be expected in each village is given by the following matrix.
\(\begin{array}{l} X \\ Y \\ Z \end{array}\left[\begin{array}{ccc} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{array}\right]\)\(\left[\begin{array}{c} 2 / 100 \\ 4 / 100 \\ 20 / 100 \end{array}\right]\) 
\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 8+12+20 \\ 6+10+15 \\ 10+16+30 \end{array}\right]\)=\(\begin{array}{l} X \\ Y \\ Z \end{array}\)\(\left[\begin{array}{c} 40 \\ 31 \\ 56 \end{array}\right]\) 
(v) (d)

(I) (d) : In factory A, number of units of types I, II and III for boys are 80, 70, 65 respectively and for girls number of units of types I, II and III are 80, 75, 90 respectively.

(ii) (a) : In factory B, number of units of types I, II and III for boys are 85, 65, 72 respectively and for girls number of units of types I, II and III are 50, 55, 80 respectively.
 
(iii) (c) : Let X be the matrix that represent the number of units of each type produced by factory A for boys, and Y be the matrix that represent the number of units of each type produced by factory B for boys.
Then, X = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[170} & 130 & 130] \end{array}\) and Y = \(\begin{array}{ccc} \text { I } & \text { II } & \text { III } \\ {[85} & 65 & 72] \end{array}\) 
Now, required matrix = X + Y = [80 70 65] + [85 65 72]
=  [165 135 137]
(iv) (a): Required matrix = [80 75 90] + [50 55 80]
= [130 130 170]
(v) (a) : Clearly,R = P+Q
\(=\left[\begin{array}{ll} 80 & 80 \\ 70 & 75 \\ 65 & 90 \end{array}\right]+\left[\begin{array}{ll} 85 & 50 \\ 65 & 55 \\ 72 & 80 \end{array}\right]=\left[\begin{array}{ll} 165 & 130 \\ 135 & 130 \\ 137 & 170 \end{array}\right]\) 
\(\therefore \quad R^{\prime}=\left[\begin{array}{lll} 165 & 135 & 137 \\ 130 & 130 & 170 \end{array}\right]\)