\(AB=\left[ \begin{matrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{matrix} \right] \left[ \begin{matrix} \frac { -4 }{ 35 } & \frac { 11 }{ 35 } & \frac { -5 }{ 35 } \\ \frac { -1 }{ 35 } & \frac { -6 }{ 35 } & \frac { 25 }{ 35 } \\ \frac { 6 }{ 35 } & \frac { 1 }{ 35 } & \frac { -10 }{ 35 } \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{matrix} \right] \frac { 1 }{ 35 } \left[ \begin{matrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{matrix} \right] \)
\(=\frac { 1 }{ 35 } \left[ \begin{matrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{matrix} \right] \)
\(=\frac { 35 }{ 35 } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =I\)
\(BA=\left[ \begin{matrix} \frac { -4 }{ 35 } & \frac { 11 }{ 35 } & \frac { -5 }{ 35 } \\ \frac { -1 }{ 35 } & \frac { -6 }{ 35 } & \frac { 25 }{ 35 } \\ \frac { 6 }{ 35 } & \frac { 1 }{ 35 } & \frac { -10 }{ 35 } \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 35 } \left[ \begin{matrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{matrix} \right] \)
\(​​=\frac { 1 }{ 35 } \left[ \begin{matrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{matrix} \right] \)
\(=\frac { 35 }{ 35 } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =I\)
Thus AB = BA = I
Therefore A and B are inverses of each other.

Given \(TR=12x+{x^2\over2}-{x^3\over 3}\)
Average Revenue \(AR={TR\over x}={12x+x^2/2-x^3/3\over x}\)
\(AR=12+{x\over2}-{x^2\over3}\)     ...(1)
Let y \(=12+{x\over2}-{x^2\over3}\)
\({dy\over dx}={1\over2}-{2x\over 3}\)
Condition for maximum is
\({dy\over dx}=0\ and\ {d^2 y\over dx^2}<0\)
\(∴\ {1\over2}-{2x\over3}=0⇒{1\over2}={2x\over 3}\)
\(x={1\over2}\times{3\over 2}={3\over 4}\)
\({d^2y\over dx^2}={-2\over3}<0\)
∴ AR is maximum at x=3/4
when x=3/4, \(AR=12+{3/4\over2}-{\left(3/4\right)^2\over 3}\)  [From )1)]
\(=12+{3\over8}-{9\over 48}=12.1875\)     ....(2)
\(MR={{dR\over dx}}={d\over dx}\left(12x+{x^2\over 2}-{x^3\over 3}\right)=12+x-x^2\)
when x = 3/4, \(MR=12+{3\over 4}-\left(3\over4\right)^2=12+{3\over4}-{9\over 16}=12.1875 \)    ..(3)
From (2) and (3), at the highest point of AR,
AR = MR = 12.1875

E1= 0	L5= 48
E2= 0+21=21	L4= 48 -11 = 37
E3 =(21 + 11) or (0 + 26)                            Whichever is maximum  =32	L3= 37 - 5 = 32
E4= (32 + 5) or (21 + 13)                                   =Whichever is maximum  = 37	L2= (37 - 13) or (32 - 11) Whichever is minimum =21
E5=37 + 11 = 48	L1=(21 - 21) or (32 - 26) Whichever is minimum = 0

Activity	Duration	EST	EFT = EST + tij	EFT = EST - tij	LFT
1-2	21	0	21	21-21=0	21
1-3	26	0	26	32-26=6	32
2-3	11	21	32	32-11=21	32
2-4	13	21	34	37-13=24	37
3-4	5	32	37	37-5=32	37
4-5	11	37	48	48-11=37	48

EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .

Let \(\bar { X } =\frac { \sum { X } }{ N } =\frac { 310 }{ 10 } \)=31
Here N=10 \(\bar { Y } =\frac { \sum { Y } }{ N } =\frac { 310 }{ 10 } \)=48

X	Y	x=X-31	y=Y-48	x2	y2	xy
22	60	-9	12	81	144	-108
24	58	-7	10	49	100	-70
26	58	-5	10	25	100	-50
28	50	-3	2	9	4	-6
30	48	-1	0	1	0	0
32	48	1	0	1	0	0
34	48	3	0	9	0	0
36	42	5	-6	25	36	-30
38	36	7	-12	49	144	-84
40	32	9	-16	81	256	-144
310	480	0	0	330	784	-492

Co-efficient correlation 
r(x,y) =\(\frac { \sum { xy } }{ \sqrt { { \sum { x } }^{ 2 } } \sqrt { { \sum { y } }^{ 2 } } } =\frac { -492 }{ \sqrt { 330 } .\sqrt { 784 } } \)
\(\Rightarrow\)\(\frac { -492 }{ 18.1659\times 28 } =\frac { -492 }{ 508.6452 } \)
\(\Rightarrow\) r=-0.9673

Given \(P(A)=\frac { 1 }{ 2 } \Rightarrow p(\bar { A } )=1-P(A)=1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 } \)
\(P(B)=\frac { 2 }{ 3 } \Rightarrow P(\bar { B } )=1-\frac { 2 }{ 3 } =\frac { 1 }{ 3 } \)
\(P(C)=\frac { 3 }{ 4 } \Rightarrow P(\bar { C } )=1-P(C)=1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
P(Only one of them hits the target)
\(=P(A\cap \bar { B } \cap \bar { C } )+P(\bar { A } \cap B\cap \bar { C } )+P(\bar { A } \cap \bar { B } \cap C)\)
\(=P(A).P(\bar { B) } .P(\bar { C) } +P(\bar { A } ).P(B).P(\bar { C } )+P(\bar { A } ).P(\bar { B } ).P(C)\)
\(=\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } \times \frac { 1 }{ 4 } +\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } \times \frac { 3 }{ 4 } \)
\(=\frac { 1 }{ 24 } +\frac { 2 }{ 24 } +\frac { 3 }{ 24 } =\frac { 6 }{ 24 } =\frac { 1 }{ 4 } \)

Let x₁cakes of the one kind and x₂ cakes of another kind are made. Let Z be the maximum number of cakes

Ingredients	x1(g)	x2(g)	Total (kg)
Flour	200	100	5
Fat	25	50	1

Thus, the mathematical formulation of the LPP is Maximize Z = x₁+ x₂
Subject to the constraints
\(200{ x }_{ 1 }+100{ x }_{ 2 }\le 5000\)
\(25{ x }_{ 1 }+50{ x }_{ 2 } \le 1000\)
\({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equations
\(200{ x }_{ 1 }+100{ x }_{ 2 }= 5000\)

\({ x }_{ 1 }\)	0	25
\({ x }_{ 2 }\)	50	0

\(25{ x }_{ 1 }+50{ x }_{ 2 }=1000\)

\({ x }_{ 1 }\)	0	25
\({ x }_{ 2 }\)	50	0

 

The feasible region is OABC and its co-ordinates are O(0, 0) A(25, 0) C(O, 20) and B is the point of intersection of the lines
200x₁ + 100x₂ = 1000 .... (1)
and 25x₁ + 50x₂ = 1000 ... (2)
Verification of B:
\((1) \Rightarrow 200{ x }_{ 1 }+100{ x }_{ 2 }=5000\)
\( (-)\quad \quad \quad (-)\quad \quad (-)\)
\( (2)\times 5\Rightarrow 50{ x }_{ 1 }+100{ x }_{ 2 }=2000\)
\(------------------\)
\(150x_{ 1 }=3000 \Rightarrow { x }_{ 1 }=20\)
\(From(2), 25(20)+50{ x }_{ 2 }=1000\Rightarrow 500+50{ x }_{ 2 }=1000 \Rightarrow 50{ x }_{ 2 }=500\)
\(\Rightarrow { x }_{ 2 }=10\)
\(\therefore B\ is\ (20,10)\)

Corner Points	Z=x1 +x2
O(0,0)	0
A(25, 0)	25
B(20, 10)	30
C(0,20)	20

Maximum of Z occurs at B(20, 10)
Hence, the solution is x₁ = 20, x₂ = 10 and Z_max = 30.

Stock = Rs.9000
FV =Rs.100
Dividend Rate = 7%
Income on 7% stock =\(\frac { \text {stock} }{ 100 } \)x FV x Rate percentage
=\(\frac { 9000 }{ 100 } \times 100\times \frac { 7 }{ 100 } \) = Rs.630    ...(1)
Cost of price of one share = 80
Sale proceeds = \(\frac { \text {stock} }{ 100 } \) x Cost Price
=\(\cfrac { 9000 }{ 100 } \)x 80 =7200
Investment = Rs.7200
Market Price = Rs.120
Dividend Rate =15%
Income =\(\frac {\text { Investment} }{ \text {Market Price } }\)x Dividend Rate
=\(\frac { 7200 }{ 120 } \)x15=Rs.900 .....(2)
Change in income = 930 - 630 = Rs.270

Let E₁, E₂ and A be defined as
E₁ = First bag is drawn
E₂ = Second bag is drawn
A = Red ball is drawn
\(\therefore\) P(E₁) = P(E₂) = \(\frac{1}{2}\)
P(A/E₁) = \(\frac{3}{7}\), P(A/E₂) = \(\frac{5}{11}\)
By Baye's theorem,
\(P({ E }_{ 1 }/A)=\frac { P({ E }_{ 1 })P(A/{ E }_{ 1 }) }{ P({ E }_{ 1 })P(A/{ E }_{ 1 })+P({ E }_{ 2 })P(A/{ E }_{ 2 }) } \)
\(=\frac { 1/2\times 3/7 }{ 1/2\times 3/7+1/2\times 5/11 } \)
\(=\frac { 3/7 }{ 3/7+5/11 } =\frac { 3/7 }{ \frac { 33+35 }{ 77 } } =\frac { 3}{7 }\times \frac { 77 }{ 68 } =\frac { 33 }{ 68 } \)

2X = 8 - 3Y
Regression equation of X on Y is
\(X=4-\frac{3}{2} Y \)
\(b_{x y}=-\frac{3}{2}<1\)
Regression equation of Y on X be
2Y = 5 - X
\(Y=\frac{5}{2}-\frac{X}{2} \)
\(b_{y x}=-\frac{1}{2}<1\)
\(\because\) both are negative, r is - ve.
\(r =-\sqrt{b_{x y} \cdot b_{y x}} \)
\(=-\sqrt{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}=\frac{-\sqrt{3}}{2} \)
\(=-\frac{1.732}{2}=-0.866\)

X	Y	x2	y2	xy
25	26	625	676	650
18	35	324	1225	630
21	48	441	2304	1008
24	28	576	784	672
27	20	729	400	540
30	36	900	1296	1080
36	25	1296	625	900
39	40	1521	1600	1560
42	43	1764	1849	1806
48	39	2304	1521	1872
\(\sum\)X = 310	\(\sum\)Y = 340	\(\sum\)X2 = 10480	\(\sum\)Y2 = 12280	\(\sum\)XY = 10718

\(r(x, y)=\frac{N \Sigma X Y-\left(\sum X\right)\left(\sum Y\right)}{\sqrt{N \Sigma X^2-(\Sigma Y)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(10718)-(310)(340)}{\sqrt{10(10480)-(310)^2} \sqrt{10(12280)-(340)^2}} \)
\(=\frac{107180-105400}{\sqrt{104800-96100} \sqrt{122800-115600}} \)
\(=\frac{1780}{\sqrt{8700 \times 7200}}=\frac{1780}{7914.54}=0.2249\)

First we have to find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant write all the inequalities of the constraints in the form of equations.
\(\therefore \) We have the lines \(x_1+x_2 \leq 6 ; x_1 \leq 4; x_2 \leq 5\)
\({ x_{ 1 } }+{ x }_{ 2 }=6\) is a line passing through the points (0,6) and (6,0).
x₁ = 4 is a line parallel to x₂ axis at a distance of 4 units.
x₂ = 5 is a line parallel to x₁ axis at a distance of 5 units.
Now we draw the graph

Corner points	Z = 3x1 + 5x2
O(0,0)	0
A(4,0)	12
B(4,2)	12 + 10 = 22
C(1,5)	3 + 25 = 28
D(5,0)	15

The optimal solution is occurs at C(1,5)
x₁ = 1, x₂ = 5, Z_max = 28
Verification
\(If \ x_1=4, \quad x_2=2 \quad \mathrm{~B}(4,2) \)
\(If \ x_2=5, \quad x_1=1 \quad C(1,5)\)

Here demand rate R = 48000
Inventory cost C₁ = 10.8% of 2.5 = \(\frac { 10.8 }{ 100 } \times 2.5=0.27\)
Ordering cost C₃ = 45
Economic order quantity q₀ = \(\sqrt { \frac { { 2C }_{ 3 }R }{ { C }_{ 1 } } } \)
\(\sqrt { \frac { 2\times 45\times 48000 }{ 0.27 } } \) = 4000 units
Number of orders per year = \(\frac { R }{ { q }_{ 0 } } \)  = \(\frac { 48000 }{ 4000 } =12\)
Time between orders t₀ = \(\frac { { q }_{ 0 } }{ R } \) = \(\frac { 1 }{ 12 } \) = 0.083 year
At EOQ, carrying cost :
=  \(\frac { { q }_{ 0 } }{ 2 } \) x C₁ \(=\frac { 4000 }{ 2 } \times 0.27\) = Rs. 540
Ordering cost = \(\frac { R }{ q_{ 0 } } \times { C }_{ 3 }\) = \(\frac { 48000 }{ 4800 } \times 45\) = Rs. 540
So at EOQ carrying cost is equal to ordering cost.

Earliest start time (EST) and latest finish time (LFT) of each activity are given in the following network.

E₁ = 0
E₂ = E₁ + t₁₂ = 0 + 8 = 8
E₃ = E₁ + t₁₃ = 0 + 4 = 4
E₄ = E₂ + t₂₄ or E₃ + t₃₄ = 8 + 10 = 18
(take E₂ + t₂₄ or E₃ + t₃₄ whichever is maximum)
E₅ = (E₂ + t₂₅ or E₄ + t₄₅) = 18 + 3 = 21
(take E₂ + t₂₅ or E₄ + t₄₅ whichever is maximum)
L₅ = 21
L₄ = L₅ – t₄₅ = 21 – 3 = 18
L₃ = L₄ – t₃₄ = 18 – 5 = 13
L₂ = L₅ – t₂₅ or L₄ – t₂₄ = 18 – 10 = 8
(take L₅ – t₂₅ or L₄ – t₂₄ whichever is minimum)
L₁ = L₂ – t₁₂ or L₃ - t₁₃ = 8 – 8 = 0
(take L₂ – t₁₂ or L₃ – t₁₃ whichever is minimum)
Here the critical path is 1-2-4-5, which is denoted by double lines.

Activity	Duration(tij)	EST	EFT=EST+tij	LST=LFT-tij	LFT
1-2	8	0	8	0	8
1-3	4	0	4	9	13
2-4	10	8	18	8	18
2-5	2	8	10	19	21
3-4	5	4	9	13	18
4-5	3	18	21	18	21

The longest duration to complete this project is 21 days.
The path connected by the critical activities is the critical path(the longest path).
Critical path is 1-2-4-5 and project completion time is 21 days.

LHS = cot x cot 2x - cot 2 x cot 3 x - cot 3x cot x
We have 3x = x + 2.x
\(cot\quad 3x=\frac { cotx\quad cot2x-1 }{ cotx+cot2x } \)   \(\left[ \therefore tan(A+B)=\frac { tanA+tanB }{ 1-tanAtanB } \right] \)
cross multiplying we get,
cot x cot 3x + cot 2x cot 3x = -1 + cot x cot 2x
\(\Rightarrow\) cot x cot 2.x - cot 2x cot 3x - cot x cot 3x = 1
Hence proved

Given y = e^a cos^-1 x    ...(1)
Differentiating with respect to 'x' we get,
\({{dy}\over{dx}}={e}^{a\ {cos}^{-1}x}.{{d}\over{dx}}\left( a\ {\cos}^{-1}x \right)={e}^{a\ {cos}^{-1}x}.\left( {{-a}\over{\sqrt{1-{x}^{2}}}} \right)\)
\(={{-ay}\over{\sqrt{1-{x}^{2}}}}\)  [ using (1) ] 
\(\Rightarrow\sqrt{1-x^2}.{{dy}\over{dx}}=-ay\)
Squaring both sides we get,
\((1-x^2).{\left( {{dy}\over{dx}} \right)}^{2}=a^2y^2\)
Differentiating again with respect to 'x' we get,
\(\left( (1-x^2).2\left( {{dy}\over{dx}} \right)\left({{d^2y}\over{dx^2}} \right) +{\left({{dy}\over{dx}} \right)}^{2}(-2x)=a^2(2y)\left( {{dy}\over{dx}} \right) \right)\)
Dividing throughout by 2 \(\left({{dy}\over{dx}} \right)\) we get,
\((1-x^2).\left({{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)=a^2 y\Rightarrow(1-x^2)\left( {{d^2y}\over{dx^2}} \right)-x\left( {{dy}\over{dx}} \right)-a^2y=0.\)
Hence proved.

Let the 3 numbers be x, y and z.
Given x + y + z = 20
2x + y - z = 23
3x+ y + z  = 46
It can be rewritten as 
\(\begin{bmatrix} 1&1&1\\2&1&-1\\3&1&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
\(\Rightarrow\)   \(A X=B \Rightarrow X=A^{-1} B\) 
Where \(A=\begin{bmatrix} 1&1& 1\\2&1&-1\\3&1&1 \end{bmatrix},X=\begin{bmatrix} x\\y\\z \end{bmatrix},B=\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
= 1 ( 1 + 1 ) - 1 ( 2 + 3 ) + 1 ( 2 - 3)
= 2 - 5 - 1 = 2 - 6 = - 4 \(\neq \) 0 
\(\therefore\) A^-1 exists.
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & -5 & -1 \\ 0 & -2 & 2 \\ -2 & 3 & -1 \end{array}\right)\)
\(\therefore{A}^{-1}={{1}\over{|A|}}adj\ A={{-1}\over{4}}\begin{bmatrix} 2&0&-2\\-5&-2&3\\-1&2&-1 \end{bmatrix}\)
\(X={A}^{-1}B={{-1}\over{4}}\begin{bmatrix}2&0&-2\\5&-2&3\\-1&2&-1 \end{bmatrix}\begin{bmatrix} 20\\23\\46 \end{bmatrix}\)
\(= \begin{bmatrix} x\\y\\z \end{bmatrix},={{-1}\over{4}}\begin{bmatrix} 40+0-92\\-100-46+138\\-20+46-46 \end{bmatrix}={{-1}\over{4}}\begin{bmatrix} -5\\-8\\-20 \end{bmatrix}\)
\(X=\begin{bmatrix} 13\\2\\5 \end{bmatrix}\)
\(x=13, y=2, z=5\)
\(\therefore\) The 3 numbers are 13, 2 and 5.

Give A = \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0\)
Applying R₂ \(\rightarrow\) R₂-R₁ and R₃ \(\rightarrow\) R₃ - R₁ 
we get A = \(\left| \begin{matrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-a & c-b \end{matrix} \right| =0\)
Expanding along C₁ we get
\(1\left| \begin{matrix} c-a & a-b \\ b-a & c-b \end{matrix} \right| =0\)
\(\Rightarrow\) (c - a) (c - b) - (b - a) (a - b) = 0
\(\Rightarrow\) c² - bc - ac + ab - (ab - b² - a² + ab) = 0
\(\Rightarrow\) a² + b² + c² - ab - bc - ca = 0
Multiplying both sides by 2 we get, 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0
\(\Rightarrow\) (a - b)² + (b - c)² + (c - a)² = 0
\(\Rightarrow\) a = b = 0, b - c = 0,c - a = 0
\(\Rightarrow\) a = b = c
\(\Rightarrow\) \(\Delta\) ABC is equilateral.
\(\therefore A=B=C=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ 2 }A+{ sin }^{ 2 }B+{ sin }^{ 2 }C=3{ sin }^{ 2 }\frac { \pi }{ 3 } =3{ \left( sin\frac { \pi }{ 3 } \right) }^{ 2 }=3{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=3\times \frac { 3 }{ 4 } =\frac { 9 }{ 4 } \)

Given A = \(\begin{bmatrix}1 &3&3 \\1 &4&3\\1&3&4 \end{bmatrix} \)
\(A_{11}=\text {Cofactor of } 1=16-9=7\)
\(A_{12}=\text {Cofactor of } 3=-(4-3)=-1\)
\(A_{13}=\text {Cofactor of } 3=3-4=-1\)
\(A_{21}=\text {Cofactor of } 1=-(12-9)=-3 \)
\(A_{22}=\text {Cofactor of } 4=4-3=1 \)
\(A_{23}=\text {Cofactor of } 3=-(3-3)=0 \)
\( A_{31}=\text {Cofactor of } 1=9-12=-3 \)
\(A_{32}=\text {Cofactor of } 3=3-3=0 \)
\(A_{33}=\text {Cofactor of } 4=4-3=1\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 7 & -1 & -1 \\ -3 & 1 & 0 \\ -3 & 0 & 1 \end{array}\right)\)
A^-1 = \(\begin{bmatrix} 7&-3&-3\\-1&1&0\\-1&0&1 \end{bmatrix}\)
\(|A| =1(16-9)-3(4-3)+3(3-4) \)
\(=7-3-3=1 \neq 0\)
\(\therefore \mathrm{A}^{-1} \text { exists }\)
\(\mathrm{A}(\operatorname{adj} \mathrm{A})=\left(\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4 \end{array}\right)\left(\begin{array}{ccc} 7 & -3 & -1 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)
\(=\left(\begin{array}{lll} 7-3-3 & -3+3+0 & -3+0+3 \\ 7-4-3 & -3+4+0 & -3+0+3 \\ 7-3-4 & -3+3+0 & -3+0+4 \end{array}\right)\)
\(=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)=|A| I\)
\(=\mathrm{A}(\operatorname{adj} \mathrm{A})=|A| I\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 7 & -3 & -3 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right)\)

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

\(\frac{x+2}{(x-1)(x+3)^2}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^2}\)
\(=\frac{A(x+3)^2+B(x-1)(x+3)+C(x-1)}{(x-1)(x+3)^2}\)
\(\Rightarrow\)  x + 2 = A ( x + 3 )² + B ( x - 1 ) ( x + 3 ) + C( x - 1 ) ....(1)
Putting x = 1 in (1) we get,
\(3=A(4)^2 \Rightarrow A=\frac{3}{16}\)
Puttingx = -3 in (1)we get
\(-1=C(-3-1)\)
\(C=\frac{1}{4}\)
Equate co-efficient of x² on both sides of (1)
\(0 =A+B \Rightarrow B=-A=\frac{-3}{16} \)
\(\frac{x+2}{(x-1)(x+3)^2} =\frac{3}{16(x-1)}-\frac{3}{16(x+3)}+\frac{1}{4(x+3)^2}\)