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

\(A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-\left|\begin{array}{ll} h & g \\ f & c \end{array}\right|\)

Area of triangle, \(\Delta=\frac{1}{2}\left|\begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|\)

As y’ = 3x² – 3, for a point of absolute maximum or minimum y’=0 ⇒ x = ± 1.
y]_x=0 = 2,
y]_x=1 = 1 – 3 + 2 = 0,
y]_x=-1 = -1 +3+ 2 = 4,
y]_x=2 = 8 – 6 + 2 = 4

As \(\frac { dx }{ dt } =2\) cm/s, x is side of equiolatral triangle.
A = \(\frac { \sqrt { 3 } }{ 4 } { x }^{ 2 }\)
\(\Rightarrow \frac { dA }{ dx } =\frac { \sqrt { 3 } }{ 2 } { x }\frac { dx }{ dt } \)
= \(\frac { \sqrt { 3 } }{ 2 } x2=\sqrt { 3 } x\)
∴\(|\frac{dA}{dx}|\)x=10 = 10\(\sqrt3\) cm²/s

As \(\left| { (AB) }^{ -1 } \right| =\frac { 1 }{ |AB| } =\frac { 1 }{ |A||B| } \)
\(\Rightarrow 8=\frac { 1 }{ 2|B| } \Rightarrow B=\frac { 1 }{ 16 } \)

Δ = 6(-1)- 1(1) = -7

As \(\begin{bmatrix} 5 & x \\ y & 0 \end{bmatrix}=\begin{bmatrix} 5 & x \\ y & 0 \end{bmatrix}\Rightarrow x=y\)

As 6 → 1 × 6, 2 × 3, 3 × 2, 6 × 1.

Let the cost of 1 pen = Rs. x, the cost of 1 bag = Rs. y, and the cost of 1 instrument box = Rs. z
According to the question, we have
5x + 3y + z = 16, 2x + Y + 3z = 19, x + 2y + 4z = 25
This system of equation can be written as AX = B,
where \(A=\left[\begin{array}{lll} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{array}\right], B=\left[\begin{array}{l} 16 \\ 19 \\ 25 \end{array}\right] \) and \(X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\) 
IAI = 5(4 - 6) - 3(8 - 3) + 1(4 - 1)
\(=-10-3(5)+3=-22 \neq 0\)
\(\therefore\)  A ^-1 exists.
Now, X = A^-1B, where \(A^{-1}=\frac{1}{|A|} \operatorname{adj} A\) 
Here, \(\operatorname{adj} A=\left[\begin{array}{ccc} -2 & -5 & 3 \\ -10 & 19 & -7 \\ 8 & -13 & -1 \end{array}\right]^{\prime}=\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad A^{-1}=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\) 
\(\therefore \quad X=\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{-22}\left[\begin{array}{ccc} -2 & -10 & 8 \\ -5 & 19 & -13 \\ 3 & -7 & -1 \end{array}\right]\left[\begin{array}{c} 16 \\ 19 \\ 25 \end{array}\right]\) 
\(=\frac{1}{-22}\left[\begin{array}{c} -32-190+200 \\ -80+361-325 \\ 48-133-25 \end{array}\right]=\frac{-1}{22}\left[\begin{array}{c} -22 \\ -44 \\ -110 \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \\ 5 \end{array}\right]\) 
\(\therefore\) x = 1, y = 2, z = 5
Hence, cost of one pen, one bag and an instrument box isRs. 1, Rs. 2 and Rs. 5 respectively.
(i) (c) : Cost of one pen is Rs. 1.
(ii) (a) : Cost of one pen. and one bag = Rs. (1 + 2) = Rs. 3
(iii) (b) : Cost of one pen and one instrument box
= Rs. (1 + 5) = Rs. 6
(iv) (c) : According to the definition of determinant, determinartt is a number associated to a square matrix.
(v) (b) : Given matrix equation is AB = AC Pre-multiplying by A^-I on both sides, we get
\(A^{-1} A B=A^{-1} A C \Rightarrow\left(A^{-1} A\right) B=\left(A^{-1} A\right) C\) 
\(\Rightarrow \quad I B=I C \quad\left(\because A A^{-1}=A^{-1} A=I\right)\) 
\(\Rightarrow B=C\) 
Since A^-1 exists only if A i's non-singular
\(\therefore\) For B = C, A should be non-singular

(i) (a) : Let x be the charges per bike per day and n be the number of bikes rented per day.
R(x) = n x x = (2000 - lOx) x = -10x² + 2000x
(ii) (b) : We have, R(x) = 2000x - 10x²
\(\Rightarrow R^{\prime}(x)=2000-20 x\) 
For R(x) to be maximum or minimum, R'(x) = 0
\(\Rightarrow 2000-20 x=0 \Rightarrow x=100\) 
Also,\(R^{\prime \prime}(x)=-20<0\) 
Thus, R(x) is maximum at x = 100
(iii) (c) : If company charge ~ 200 or more, they will not rent any bike. Therefore, revenue collected by him will be zero.
(iv) (c) : If x = 105, number of bikes rented per day is given by
n = 2000 - 10 x 105 = 950
(v) (d) : At x = 100, R(x) is maximum
\(\therefore\) Maximum revenue = R(100)
= -10(100)² + 2000(100) = Rs. 1,00,000

(i) (d) : Let be the area of the triangle then,

\(\begin{equation} =\frac{1}{2}|-2(-6-5)-6(3-1)+1(15+6)| \end{equation}\) [Expanding along R_1]
\(\begin{equation} \Rightarrow \quad \Delta=\frac{1}{2}|43-12|=15.5 \mathrm{sq} . \text { units } \end{equation}\) 
(ii) (d) : The given points are collinear.
\(\begin{equation} \therefore \frac{1}{2}\left|\begin{array}{ccc} 2 & -3 & 1 \\ k & -1 & 1 \\ 0 & 4 & 1 \end{array}\right|=0 \end{equation}\) 
Expanding along R₁',we get
\(\begin{equation} 2(-1-4)+3(k)+1(4 k)=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 7 k-10=0 \Rightarrow k=\frac{10}{7} \Rightarrow 4 k=\frac{40}{7} \end{equation}\) 
(iii) (a) : Area of \(\begin{equation} \Delta A B C=3 \text { sq. units } \end{equation}\) [Given]
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 3 \Rightarrow\left|\begin{array}{lll} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{array}\right|=\pm 6 \end{equation}\) 
\(\begin{equation} \Rightarrow 1(0-0)-3(0-k)+1(0-0)=\pm 6 \end{equation}\) 
\(\begin{equation} \Rightarrow 3 k=\pm 6 \Rightarrow k=\pm 2 . \end{equation}\) 
(iv) (a) : Let Q(x, y) be any point on the line joining
A(1, 2) and B(3, 6). Then, area of \(\begin{equation} \Delta A B Q=0 \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{1}{2}\left|\begin{array}{lll} 1 & 2 & 1 \\ 3 & 6 & 1 \\ x & y & 1 \end{array}\right|=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 1(6-y)-2(3-x)+1(3 y-6 x)=0 \end{equation}\) 
\(\begin{equation} \Rightarrow 6-y-6+2 x+3 y-6 x=0 \end{equation}\) 
\(\begin{equation} \Rightarrow-4 x=-2 y \Rightarrow 2 x=y \end{equation}\)
(v) (c) : Area of ΔABC is given by
\(\begin{equation} \frac{1}{2}\left|\begin{array}{rrr} 11 & 7 & 1 \\ 5 & 5 & 1 \\ -1 & 3 & 1 \end{array}\right|=\frac{1}{2}[11(5-3)-7(5+1)+1(15+5)] \end{equation}\) 
\(\begin{equation} =\frac{1}{2}[22-42+20]=0 \end{equation}\) 
\(\therefore\) Points are collinear

(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)