TU Solution

Thus, line (ii) is passing through the points (0, 14)
and (7, 0).
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 100 which is true.
From inequality (ii), we get 0 ≤ 140 which is true.
As, we have given x ≥ 0, and y ≥ 0, the feasible
region lies on first quadrant only.
Plotting above points in the graph, we have

Calculation of Maximum Value

Vertices Z = 20x + 30y Remark

O(0, 0) Z = 20×0 + 30×0 = 0

A(7, 0) Z = 20×7 + 30×0 = 140

B(6, 2) Z = 20×6 + 30×2 = 180 Maximum

C(0, 5) Z = 20×0 + 30×5 = 150

Therefore, the maximum value is 180 at B(6, 2).

QN.5) (2075) Food X contains 6 units of vitamin A per gm, 7 units of vitamin B per gm & cost
Rs.5 per gm. Food Y contains 8 units of vitamin A per gm, 12 units of vitamin B per gm and cost
Rs.18 per gm. The daily minimum requirement of vitamin A and B are 100 units and 120 units
respectively. Formulate the given problem in a LLP model with the objective of minimizing the
cost.
Solution: Let food X and Y produce x and y units respectively. Then
Linear Programming Problem model of given problem
Minimize: C = 5x + 18y
subject to the constraints:
6x + 8y ≥ 100
7x + 12y ≥ 120
and x ≥ 0, y ≥ 0
Let associated equations of the given constraints are
6x + 8y = 100 ………….. (i)
7x + 12y = 120 ……….. (ii)

250

From (i), we get

If x = 0, y = 12.5

If y = 3, x = 14

Thus, the line (i) passes through the points (0, 12.5) and (14, 3).

From (ii)

If x = 0, y = 12, If y = 3, x = 12

Thus, line (ii) is passing through the points (0, 12) and (12, 3).

To identity feasible region, we use

test point (0, 0).

From inequality (i), we get 0 ≥ 100

which is false.

From inequality (ii), we get 0 ≥ 120

which is false.

As, we have given x ≥ 0, and y ≥ 0,

the feasible region lies on first

quadrant only.

Plotting above points in the graph, we

have

Calculation of Minimum Value

Vertices C = 5x + 18y Remark

C(15, 1.25) C = 5×15 + 18×1.25 = 116.25 Minimum

D(0, 12.5) C = 5×0 + 18×12.5 = 225

E(0, 10) C = 5×0 + 18×10 = 180

Therefore, minimum value of C is 116.5 at C(15, 1.25).
QN. 6) (2074) Solve the following Linear Programming Problem by Graphical Method
Objective function Zmax = 600x + 300y
subject to the constraints:
x + 2y ≤ 8
x+y≥2
5x + 3y ≤ 15
and x ≥ 0, y ≥ 0

251

Solution:
Let associated equations of the given constraints are
x + 2y = 8 …………. (i)
x + y = 2 ………….. (ii)
5x + 3y = 15 ………. (iii)
From (i), we get
If x = 0, y = 4
If y = 0, x = 8
Thus, the line (i) passes through the points (0, 4) and (8, 0).
From (ii)
If x = 0, y = 2
If y = 0, x = 2
Thus, line (ii) is passing through the points (0, 2) and (3, 0).
From (iii)
If x = 0, y = 5
If y = 0, x = 3
Thus, line (iii) is passing through the points (0, 5) and (3, 0).
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 8 which is true.
From inequality (ii), we get 0 ≥ 2
which is false.
From inequality (iii), we get 0 ≤ 15
which is true.
As, we have given x ≥ 0, and y ≥ 0,
the feasible region lies on first
quadrant only.
Plotting above points in the graph,
we have

Vertices Calculation of Maximum Value Remark
Z = 600x + 300y

252

A(2, 0) Z = 600×2 + 300×0 = 1200

B(3, 0) Z = 600×3 + 300×0 = 1800 Maximum

D(0.86, 3.57) Z = 600×0.86 + 300×3.57 = 1587

F(0,4) Z = 600×0 + 300×4 = 1200

G(0, 2) Z = 600×0 + 300×2 = 600

Therefore, maximum value of Z is 1800 at B(3, 0).
QN. 7) (2074, old) Solve the following Linear Programming Problem using Graphical Method
Maximize Z = 30x + 50y
subject to the constraints:
x + y ≤ 30
x + 2y ≤ 40
and x ≥ 0, y ≥ 0
Solution:
Let associated equations of the given constraints are
x + y = 30 …………. (i)
x + 2y = 40 ………….. (ii)
From (i), we get
If x = 0, y = 30
If y = 0, x = 30
Thus, the line (i) passes through the points (0, 30) and (30, 0).
From (ii)
If x = 0, y = 20
If y = 0, x = 40
Thus, line (ii) is passing through the points (0, 20) and (40, 0).

253

To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 30 which is
true.
From inequality (ii), we get 0 ≤ 40 which is
true.
As, we have given x ≥ 0, and y ≥ 0, the
feasible region lies on first quadrant only.
Plotting above points in the graph, we have

Vertices Calculation of Maximum Value Remark
O(0, 0) Z = 30x + 50y Maximum
A(30, 0) Z = 30×0 + 50×0 = 0
C(20, 10) Z = 30×30 + 50×0 = 900
D(0,20) Z = 30×20 + 50×10 = 1100
Z = 30×0 + 50×20 = 1000

Therefore, maximum value of Z is 1100 at C(20, 10).

QN. 8) (2073) Solve the following Linear Programming Problem graphically
Objective function min Z = 3x + 2y
subject to the constraints:
5x + y ≥ 10
2x + 4y ≥ 12
x+y≤6
and x ≥ 0, y ≥ 0
Solution:
Let associated equations of the given constraints are
5x + y = 10 …………. (i)
2x + 4y = 12 ………….. (ii)
x + y = 6 ………. (iii)

254

From (i), we get
If x = 0, y = 10
If y = 0, x = 2
Thus, the line (i) passes through the points (0, 10) and (2, 0).
From (ii)
If x = 0, y = 3
If y = 0, x = 6
Thus, line (ii) is passing through the points (0, 3) and (6, 0).
From (iii)
If x = 0, y = 6
If y = 0, x = 6
Thus, line (iii) is passing through the points (0, 6) and (6, 0).
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≥ 10 which is false.
From inequality (ii), we get 0 ≥ 12 which is false.
From inequality (iii), we get 0 ≤ 6 which is true.
As, we have given x ≥ 0, and y ≥ 0, the feasible region lies on first quadrant only.
Plotting above points in the graph, we have

255

Calculation of Maximum Value

Vertices Z = 3x + 2y Remark
Minimum
B(6, 0) Z = 3×6 + 2×0 = 18

C(1, 5) Z = 3×1 + 2×5 = 13

G(1.56, 2.22) Z = 3×1.56 + 2×2.22 = 9.12

Therefore, minimum value of Z is 9.12 at G(1.56, 2.22).
QN.9) (2073, old) One unit of food X contains 2 grams of vitamin A and 2 grams of vitamin B
and one unit of food Y contains 1 unit of vitamin A and 3 units of vitamin B. A man wants to
produce as many food as possible when 20 grams of vitamin A and 24 grams of vitamin B are
available. If the profit of selling food X and food Y are Rs.50 and Rs. 30 respectively. Formulate
the given problem in mathematical form and solve it graphically in order to maximize the profit.
Solution: Let food X and Y produce x and y units respectively. Then the mathematical form of
given problem is
Maximize: Z = 50x + 30y
subject to the constraints:
2x + y ≤ 20

256

2x + 3y ≤ 24

and x ≥ 0, y ≥ 0

Let associated equations of the given constraints are

2x + y = 20 ………….. (i)

2x + 3y = 24 ……….. (ii)

From (i), we get

If x = 0, y = 20

If y = 0, x = 10

Thus, the line (i) passes through the points (0, 20) and (10, 0).

From (ii)

If x = 0, y = 8

If y = 0, x = 12

Thus, line (ii) is passing through the points (0, 8)

and (12, 0).

To identity feasible region, we use test point (0, 0).

From inequality (i), we get 0 ≤ 20 which is true.

From inequality (ii), we get 0 ≤ 24 which is true.

As, we have given x ≥ 0, and y ≥ 0, the feasible

region lies on first quadrant only.

Plotting above points in the graph, we have

Calculation of Maximum Value

Vertices Z = 50x + 30y Remark

O(0, 0) Z = 50×0 + 30×0 = 0

A(10, 0) Z = 50×10 + 30×0 = 500

C(9, 2) Z = 50×9 + 30×2 = 510 Maximum

E(0, 8) Z = 50×0 + 30×8 = 240

Therefore, the maximum value Z is 510 at x = 9 and y = 2.
QN. 10) (2072) Use the graphical method, solve the following linear programming problems
Maximize Z = 2x + 3y
subject to the constraints:
3x + 5y ≤ 45

257

6x + 4y ≤ 48

and x ≥ 0, y ≥ 0

Solution:

Let associated equations of the given constraints are

3x + 5y = 45 …………. (i)

6x + 4y = 48 ………….. (ii)

From (i), we get

If x = 0, y = 15

If y = 0, x = 9

Thus, the line (i) passes through the points (0, 15) and (9, 0).

From (ii)

If x = 0, y = 8

If y = 0, x = 12

Thus, line (ii) is passing through the points (0, 8)

and (12, 0).

To identity feasible region, we use test point (0, 0).

From inequality (i), we get 0 ≤ 45 which is true.

From inequality (ii), we get 0 ≤ 48 which is true.

As, we have given x ≥ 0, and y ≥ 0, the feasible

region lies on first quadrant only.

Plotting above points in the graph, we have

Calculation of Maximum Value

Vertices Z = 2x + 3y Remark

O(0, 0) Z = 2×0 + 3×0 = 0

A(8, 0) Z = 2×8 + 3×0 = 16

C(3.33, 7) Z = 2×3.33 + 3×7 = 27.66 Maximum

E(0, 9) Z = 20×0 + 3×9 = 27

Therefore, the maximum value of Z is 27.66 at x = 3.33 and y = 7.

QN. 11) (2072, old) A dealer wishes to purchase number of fans and swing machines. He has only

Rs.5,760 to invest and has spaces for at most 20 items. A fan costs him Rs.360 and a swing

machine Rs.240. his expectation is that he can sale a fan at a profit of Rs.22 and swing machine at

258

a profit of Rs.18. How should he invest his money in order to maximize his profit by using graphic

method?

Solution: Let the dealer purchase number of x fans and y swing machines. Now, the mathematical

form of given problem is

Maximize: P = 22x + 18y

subject to the constraints 360x + 240y ≤ 5760, x + y ≤ 20, x & y ≥ 0.

Let associated equations of the given constraints are

360x + 240y = 5760 …………. (i)

x + y = 20 ………….. (ii)

From (i), we get

If x = 0, y = 24

If y = 0, x = 16

Thus, the line (i) passes through the points (0, 24) and (16, 0).

From (ii)

If x = 0, y = 20

If y = 0, x = 20

Thus, line (ii) is passing through the points

(0, 20) and (20, 0).

To identity feasible region, we use test

point (0, 0).

From inequality (i), we get 0 ≤ 5760 which

is true.

From inequality (ii), we get 0 ≤ 20 which is

true.

As, we have given x ≥ 0, and y ≥ 0, the

feasible region lies on first quadrant only.

Plotting above points in the graph, we have

Calculation of

Maximum Value

Vertices P = 22x + 18y Remark

O(0, 0) Z = 22×0 + 18×0 = 0

A(16, 0) Z = 22×16 + 18×0 = 352

259

C(8, 12) Z = 22×8 + 18×12 = 392 Maximum

E(0, 20) Z = 22×0 + 18×20 = 360

Therefore, the maximum value of P is 392 at x = 8 and y = 12.

QN. 12) (2071) Find the graphically the minimum value of Z = 40x + 20y

subject to constraints

x + 2y ≥ 2, 3x + y ≥ 3, 4x + 3y ≥ 6; x ≥ 0, y ≥ 0.

Solution: Let associated equations of the given constraints are

x + 2y = 2 …………. (i)

3x + y = 3 ………… (ii)

4x + 3y = 6 ………. (iii)

From (i), we get

If x = 0, y = 1

If y = 0, x = 2

Thus, the line (i) passes through the points (0, 1) and (2, 0).

From (ii)

If x = 0, y = 3

If y = 0, x = 1

Thus, line (ii) is passing through the points (0, 3) and (1, 0).

From (iii)

If x = -3, y = 6

If y = -2, x = 3

Thus, line (iii) is passing through the points (-3, 6) and (-2, 3).

To identity feasible region, we use test point (0, 0).

From inequality (i), we get 0 ≥ 2 which is false.

From inequality (ii), we get 0 ≥ 3 which is false.

From inequality (iii), we get 0 ≥ 6 which is false.

As, we have given x ≥ 0, and y ≥ 0, the feasible region lies on first quadrant only.

Plotting above points in the graph, we have

Calculation of Minimum Value

260

Vertices Z = 40x + 20y Remark
C(2, 0) Z = 40×2 + 20×0 = 80 Minimum
D(1.2, 0.4) Z = 40×1.2 + 20×0.4 = 56
F(0.6, 1.2) Z = 40×0.6 + 20×1.2 = 48
E(0, 3) Z = 22×0 + 18×3 = 54

Therefore, the minimum value of Z is 48 at x = 0.6 and y = 1.2.

QN. 13) (2070) Solve the following Linear Programming problem by using graphical method
Maximize the profit = Rs.300A + Rs.500B
subject to constraints
9A + 6B ≤ 54, 3A ≤ 18, B ≥ 6, where A, B ≥ 0, non-negative conditions.
Solution: Let associated equations of the given constraints are
9A + 6B = 54 …………. (i)
3A = 18 ………… (ii)
B = 6 ………. (iii)
From (i), we get
If A = 0, B = 9
If B = 0, A = 6
Thus, the line (i) passes through the points (0, 9) and (6, 0).
From (ii), we have A = 6
Thus, line (ii) is passes through (6, 0) and parallel to the y-axis.

261

From (iii) B = 6
Thus, line (iii) is passing through the points (0, 6) and
parallel to the x-axis.
To identity feasible region, we use test point (0, 0).
From inequality (i), we get 0 ≤ 54 which is true.
From inequality (ii), we get 0 ≤ 3 which is true.
From inequality (iii), we get 0 ≥ 6 which is false.
As, we have given A ≥ 0, and B ≥ 0, the feasible
region lies on first quadrant only.
Plotting above points in the graph, we have

Calculation of Maximum Profit

Vertices Profit (P) = Rs.300A + Rs.500B Remark
C(2, 6) P = Rs.300×2 + Rs.500×6 = Rs.3600
D(0, 6) P = Rs.300×0 + Rs.500×6 = Rs.3000

E(0, 9) P = Rs.300×0 + Rs.500×9 = Rs.4500 Maximum

Therefore, the maximum profit isRs.4500 at A = 0 and B = 9.

262

Chapter - 13

Determinant

Brief Question Answer

156
QN. 1) (2078) Find the determinant [2 3 7]

356
156
Now, [2 3 7] = 1(36 − 35) − 5(12 − 21) + 6(10 − 9) = 1 + 45 + 6 = 52
356

QN.2) (2077, Regular) An Aeroplan has 150 seats for passengers. Every passenger will to

pay business class fare is allowed to carry 40 Kgs of baggage each. But economic class

passengers are restricted to carry only 20 Kgs of baggage each. If 4000 Kgs of baggage

can be carried, find the number of passengers of each kind be using determinant method.

Solution: Let number of passengers in business class be x and that of economic class be y.

Then, we have x + y = 150

40x + 20y = 4000

Now, = |410 210| = 20 − 40 = −20

= |4105000 210| = 3000 − 4000 = −1000

= |410 4105000| = 4000 − 6000 = −2000

∴ = 1 = − 1000 = 50, = 2 = − 2000 = 100
−20 −20

147
QN.3) (2077, Back) Find the value of determinant [2 5 8]

369

147
Now, |2 5 8| = 1(45 − 48) − 4(18 − 24) + 7(12 − 15) = −3 + 16 − 21 = −8

369
347

QN.4) (2077, Regular) Find the value of determinant [2 7 8]
532

Solution: 7
34 8| = 3(14 − 24) − 4(4 − 40) + 7(6 − 35) = −30 + 144 − 203 = −98
2
Now, |2 7
53

263

Descriptive Question Answer

QN.5) (2077, New) Solving following equations by using determinate or matrix method

2x + 2y + z = 5

4x + 2y + 3z = 10

x + 3y + 4z = 7

Solution: Now,

221
= |4 2 3| = 2(8 − 9) − 2(16 − 3) + 1(12 − 2) = −2 − 26 + 10 = −18

134

5 21
= |10 2 3| = 5(8 − 9) − 2(40 − 21) + 1(30 − 14) = −5 − 38 + 16 = −27

7 34

251
= |4 10 3| = 2(40 − 21) − 5(16 − 3) + 1(28 − 10) = 38 − 65 + 18 = −9

174

22 5
= |4 2 10| = 2(14 − 30) − 2(28 − 10) + 5(12 − 2) = −32 − 36 + 50 = −18

13 7

∴ = = − −27 = 3 , = − −9 = 1 , = −18 = 1
−18 2 −18 2 −18

QN.5) (2077, New) By using determinant method. Solve the following linear equations:

3x + 4y + 5z = 18

2x – y – 8z = 13
5x – 2y + 7z = 20

34 5
Solution: Now, = |2 −1 −8| = 3(−7 − 16) − 4(14 + 40) + 5(−4 + 5)

5 −2 7
= −69 − 216 + 5 = −280

18 4 5
= |13 −1 −8| = 18(−7 − 16) − 4(91 + 160) + 5(−26 + 20) = −414 − 1004 − 30

20 −2 7
= −1448

3 18 5
= |2 13 −8| = 3(91 + 160) − 18(14 + 40) + 5(40 − 65)

5 20 7
= 753 − 972 − 125 = −344

264

3 4 18
= |2 −1 13| = 3(−20 + 26) − 4(40 − 65) + 18(−4 + 5)

5 −2 20
= 18 + 100 + 18 = 136

∴ = = − 1448 = 181 , = − −344 = 43 , = 136 = −17
−280 35 −280 35 −280 35

QN.7) (2077, old) Solve the following linear equations by using determinant or matrix

method:

2x + 2y + 2z = 6

4x + 2y +3z = 10

x + 3y + 4z = 7

222
Solution: Now, = |4 2 3| = 2(8 − 9) − 2(16 − 3) + 2(12 − 2)

134

= −2 − 26 + 20 = −8

6 22
= |10 2 3| = 6(8 − 9) − 2(40 − 21) + 2(30 − 14)

7 34

= −6 − 38 + 32 = −12

262
= |4 10 3| = 2(40 − 21) − 6(16 − 3) + 2(28 − 10)

174

= 38 − 78 + 36 = −4

22 6
= |4 2 10| = 2(14 − 30) − 2(28 − 10) + 6(12 − 2)

13 7

= −32 − 36 + 60 = −8

∴ = = −12 = 3 , = − −4 = 1 , = −8 = 1
−8 2 −8 2 −8

QN. 8) (2076). Prove the following determinant by using properties:
+ + 2
| + + 2 | = 2( + + )3
+ + 2

+ + 2
Proof: LHS = | + + 2 |
+ + 2

265

+ + 2
= | + + 2 | , 1 → 1 − 2
+ + 2

+ + −( + + ) 0
=| + + 2 |
+ + 2

1 −1 0
= ( + + ) | + + 2 |
+ + 2

1 −1 0
= ( + + ) | + + 2 | , 3 → 3 − 2
0 −( + + ) + +

1 −1 0
= ( + + )2 | + + 2 |

0 −1 1
0 −1 0

= ( + + )2 | + 2 + 2 + + 2 | 1 → 1 + 2
−1 −1 1

= ( + + )2 1( + 2 + 2 + )

= ( + + )2(2 + 2 + 2 )

= ( + + )22( + + )

= 2( + + )3
= RHS

Hence proved.

QN. 9) (2075) Hari sells 7 shares of P & 9 shares of Q thus increasing his cash by Rs. 70.
Gopal sells 9 shares of P & buys 14 shares of Q, decreasing his cash by Rs.80. Find the price of
P and Q by using Cramer's rule.

Solution: Let the cost of each share of P and Q be x and y respectively.
Here given
7x - 9y = 70 …………..(i)
9x - 14y = -80 ………..(ii)
Now,
= |97 −−194| = −7 × 14 + 9 × 9 = −98 + 81 = −17

= |−7800 −−194| = −70 × 14 + 80 × 9 = −980 − 720 = −1700

266

= |79 −7800| = −7 × 80 − 9 × 70 = 560 − 630 = −1190

∴ = = − 1700 = 100, = = − −1190 = 70
17 17
Hence, the price per share of A and B Rs. 100 and Rs. 70 respectively.

QN.10) (2074). Prove the following by using the properties of determinant:
2 + 1

| 2 + 1 | = 1 + 2 + 2 + 2
2 + 1

Proof:
2 + 1 2 + 1 |
2 + 1
LHS: |



1 ( 2 + 1) 2 2
| 2 ( 2 + 1) 2 |
= ( 2 + 1)
2 2

( 2 + 1) 2 2
| 2 ( 2 + 1) 2 |
= ( 2 + 1)
2 2

2 + 1 + 2 + 2 2 + 1 + 2 + 2 2 + 1 + 2 + 2
= | 2 2 + 1 2 | , [ 1 → 1 + 2 + 3]
2
2 2 + 1

1 1 1
= (1 + 2 + 2 + 2) | 2 2 + 1 2 |
2 + 1
2 2

0 1 1
= (1 + 2 + 2 + 2) |−1 2 + 1 2 |, [ 1 → 1 − 2]
2 + 1
0 2

= (1 + 2 + 2 + 2)[−(−1)( 2 + 1 − 2)]

= (1 + 2 + 2 + 2)[( 2 + 1 − 2)]

= (1 + 2 + 2 + 2)

= RHS
Hence proved.

267

QN.11) (2074, Old). Prove the following determinant:


| 2 2 2| = ( − )( − )( − )
3 3 3
Proof: = | 2 2 2|

3 3 3

111
= | | [Taking common abc]

2 2 2

0 01 [Applying, C2, → C2 − C3]
= | − − | C1 → C1 − C2
2 − 2 2 − 2 2

0 01
= ( − )( − ) | 1 1 |
+ + 2

= ( − )( − ) [1( + − − )]

= ( − )( − )( − )

= RHS

Hence proved.

QN.12) (2073) Solve the following by using the properties of determinant:

7 + 7 − 7 −
|7 − 7 + 7 − | = 0
7 − 7 − 7 +
Solution:

7 + 7 − 7 −
|7 − 7 + 7 − | = 0
7 − 7 − 7 +

[Applying, C1 → C1 + C2 + C3]
21 − 7 − 7 −

⇒ |21 − 7 + 7 − | = 0
21 − 7 − 7 +

1 7 − 7 −
⇒ (21 − ) |1 7 + 7 − | = 0

1 7 − 7 +

[Applying, R1 → R1 − R2]
0 −2 0

⇒ (21 − ) |1 7 + 7 − | = 0
1 7 − 7 +

⇒ (21 − )[2 (7 + − 7 + )] = 0

268

⇒ (21 − )2 2 = 0
Eirher 2x2 = 0 ⇒ x = 0
or 21 − x = 0 ⇒ x = 21

QN.13) (2073, Old) Prove the following by using the properties of determinant:

+ + 2
| + 2 + | = 9 2( + )
+ + 2

+ + 2
= | + 2 + |
+ + 2

3 + 3 3 + 3 3 + 3
= | + 2 + | [Appying, R1 → R1 + R2 + R3]
+ + 2

1 11
= 3( + ) | + 2 + |

+ + 2

01 0
= 3( + ) |2 | [Applying, C1 → C1 − C2, C3 → C3 − C2]
− + 2 −2

010
= 3 2( + ) | 2 1 |

−1 + 2 −2

= 3 2( + )[−1(−4 + 1)]

= 9 2( + )

= RHS

Hence proved.

QN.14) (2072) By using the properties of determinant, prove that

+ + − −
| − + + | = 2( + )( + )( + )
− − + +

Proof:

+ + − −
: | − + + − |
− − + +

+ − −( + )
= | + + + + | [Applying, C1 → C1 + C2, C3 → C3 + C2]
−( + ) − +

269

1 − −1
= ( + )( + ) | 1 + + 1 |
−1 − 1

0 − −1
= ( + )( + ) |2 + + 1 | [Applying, C1 → C1 + C2]
0 − 1

= −2( + )( + )(− − )

= 2( + )( + )( + )

= RHS

QN.15) (2072) Show that
2 3
| 2 3| = ( − )( − )( − )
2 3
Solution: Same as QN.7

QN. 16) (2072, Old) The price of three commodities X, Y, and Z are rupees x, y, and z per unit

respectively. A purchase 4 units of Z and sells 3 units of X and 5 units of Y, B purchase 3 units

of Y and sells 2 units of X and I unit of Z. C purchases 1 unit of X and sells 4 units of Y and 6

units of Z. In the process A, B and C earn Rs. 6000, Rs. 5000, and Rs.13000 respectively. Find
the price per unit of three commodities by using Cramer’s rule.

Solution:

Represent the given information in the table:
X Y Z Profit

A 3 5 -4 6000

B2 -3 1 0

C -1 4 6 13000

We can write the above information as
3x + 5y - 4z =6000 ……………… (i)
2x - 3y +z = 5000 …………………….. (ii)
- x +4y + 6z = 13000 ………… (iii)

3 5 −4
Now, = | 2 −3 1 |

−1 4 6
= 3(−18 − 4) − 5(12 + 1) − 4(8 − 3)
= −66 − 65 − 20 = −151

270

6000 5 −4
= | 5000 −3 1 |

13000 4 6
= 6000(−18 − 4) − 5(30000 − 13000) − 4(20000 + 39000)

= −132000 − 85000 − 236000
= −453000

3 6000 −4
= | 2 5000 1 |

−1 13000 6
= 3(30000 – 13000) – 6000(12 + 1) – 4(26000 + 5000)
= 51000 – 78000 – 124000
= - 151000

3 5 6000
= | 2 −3 5000 |

−1 4 13000
= 3(-39000 – 20000) – 5(26000 + 5000) – 6000(8 - 3)
= -177000 – 155000 – 30000
= - 302000

= = − 453000 = 3000, = = − 151000 = 1000, = = − 302000 = 2000
−151 −151 −151

The price per unit of X is Rs.3000, per unit of Y is Rs.1000, and per unit of Z is Rs.2000.

QN. 17) (2072, Old) Prove by using the properties of determinants that

+ + 2
| + + 2 | = 2( + + )3
+ + 2

Proof:
+ + 2 + + 2 |
+ + 2
LHS: |


+ + 2
= | + + 2 | , 1 → 1 − 2
+ + 2

+ + −( + + ) 0
= | + + 2 |
+ + 2

271

1 −1 0
= ( + + ) | + + 2 |
+ + 2

1 −1 0
= ( + + ) | + + 2 | , 3 → 3 − 2
0 −( + + ) + +

1 −1 0
= ( + + )2 | + + 2 |

0 −1 1
0 −1 0

= ( + + )2 | + 2 + 2 + + 2 | 1 → 1 + 2
−1 −1 1

= ( + + )2 1( + 2 + 2 + )

= ( + + )2(2 + 2 + 2 )

= ( + + )22( + + )

= 2( + + )3
= RHS

Hence proved.
QN. 18) (2071) Find the numerical value of the following determinant by using the

properties.
+
| + |
+

Solution:
+ |
| +
+

+
= | + | [ 1 → 1 + 2 + 3]
+

2( + + ) + + + +
= | + |
+

2 11
= ( + + ) | + |

+

272

0 01
= ( + + ) | − − | [Applying C1 → C1 − 2C2, C2 → C2 − C3]
− −

= ( + + )( − − + 2 − + 2 + − )

= ( + + )( − )2

QN. 19) (2071, Old) Prove the following by using the properties of determinants

111
| | = ( − )( − )( − )( + + )
2 2 2

Proof:

111
: | |

3 3 3

0 0 1 [Applying, C3]
= | − − | C1 → C1 − C2, C2 → C2 −
3 − 3 3
3 − 3

= ( − )( − ) | 0 01
1 1 |
2 + + 2 2 + + 2 3

= ( − )( − )( 2 + + 2 − 2 − − 2)

= ( − )( − )[( ( − ) + ( − )( + )]

= ( − )( − )( − )( + + )

= RHS

Hence proved.

QN. 20) (2070) Prove the following by using the properties of determinants
111

| | = ( − )( − )( − )( + + )
2 2 2
Proof: See question no. 15

QN.21 (2062) Prove the following by using the properties of determinant:
2 + 1

| 2 + 1 | = 1 + 2 + 2 + 2
2 + 1

Proof:
2 + 1 2 + 1 |
2 + 1
LHS: |



273

1 ( 2 + 1) 2 2
| 2 ( 2 + 1) 2 |
= ( 2 + 1)
2 2

( 2 + 1) 2 2
| 2 ( 2 + 1) 2 |
= ( 2 + 1)
2 2

2 + 1 + 2 + 2 2 + 1 + 2 + 2 2 + 1 + 2 + 2
= | 2 2 + 1 2 | , [ 1 → 1 + 2 + 3]
2
2 2 + 1

1 1 1
= (1 + 2 + 2 + 2) | 2 2 + 1 2 |
2 + 1
2 2

0 1 1
= (1 + 2 + 2 + 2) |−1 2 + 1 2 |, [ 1 → 1 − 2]
2 + 1
0 2

= (1 + 2 + 2 + 2)[−(−1)( 2 + 1 − 2)]

= (1 + 2 + 2 + 2)[( 2 + 1 − 2)]

= (1 + 2 + 2 + 2)

RHS

Hence proved.
QN. 22) (2066) Solve the following problem by using Cramer's rule.

4 1 + 2 2 + 3 = 15
1 + 4 2 + 3 3 = 28
6 1 − 2 − 3 = −2

42 1
Now, = |1 4 3 | = 4(−4 + 3) − 2(−1 − 18) + 1(−1 − 24) = −4 + 38 − 25 = 9

6 −1 −1
15 2 1
= |28 4 3 | = 15(−4 + 3) − 2(−28 + 6) + 1(−28 + 8) = −15 + 44 − 20 = 9
−2 −1 −1
4 15 1
= |1 28 3 | = 4(−28 + 6) − 15(−1 − 18) + 1(−2 − 168) = −88 + 285 − 170 = 27
6 −2 −1

274

4 2 15
= |1 4 28| = 4(−8 + 28) − 2(−2 − 168) + 15(−1 − 24) = 80 + 340 − 375 = 45

6 −1 −2

∴ = = 9 = 1, = = 27 = 3, = = 45 = 5
9 9 9

+
QN. 23) (2056) Prove that | + | = 4
+

+
LHS: | + |
+

−2 −2 0
= | + | , 1 → 1 − 2 − 3

+

1 10
=-2b | + | ,

+

0 10
=-2b | | , 1 → 1 − 2

− +

= −2 (−1)( + )
= 4
RHS

Hence proved.

QN. 22) (2064, Old) Prove the following by using the properties of determinant:
2 ( + )2 2
| 2 2 ( + )2| = 2 ( + + )2
( + )2 2 2

Solution: ( + )2 2
2 2 ( + )2|
2
LHS:| 2 2
( + )2

2 − 2 ( + )2 − 2 2 − ( + )2
= | 2 2 ( + )2 |[ ∵ 1 → 1 − 2, 3 → 3 − 2]

( + )2 − 2 2 − 2 2 − ( + )2

275

0 ( + − )( + + ) ( − − )( + + )
= | 2 2 ( + )2 |
( + − )( + + ) 0 ( − − )( + + )

0 ( + − ) ( − − )
= ( + + )2 | 2 2 ( + )2 |
( + − ) 0 ( − − )

0 ( + − ) −2
= ( + + )2 | 2 2 2 | , 3 → 3 − 2 − 1
( + − ) 0 −2

0 ( + − ) −
= 2( + + )2 | 2 2 |
( + − ) 0 −

2 ( + ) 0
2
= 2( + + )2 | | , 1 → 1 + 2 , 3 → 3 + 2
2
2

( + ) 0

= 2( + + )2(− )( − − − 2 − )

= 2( + + )2( )( 2 + + )

= 2 ( + + )3

RHS
Hence proved.
QN. 24) (2020) prove that following by using properties.

− 2
| − 2 | = 4 2 2 2
− 2


Solution:
− 2
− 2
LHS: | − 2|


−
− − |
= |

276

0
= |20 − − |


= 2 2( + )
= 4 2 2 2

RHS
Hence proved.

277

Chapter - 14

Matrix

Brief Question Answer

78 65
QN. 1) (2078) Find 7(A – B) where A = [9 10] and B = [3 2]
25 75

7 8 6 5 7+6 8+5 13 13
Solution: Now, 7(A – B) = [9 10] [3 2] = [9 + 3 10 + 2] = [12 12]
2 5 7 5 2+7 5+5 9 10

QN. 2) (2077, New) Show that [35 −−21] [25 −−13] are inverse of each other.

Solution: Now, [53 −−12] [52 −−13] = [106 − 5 −3 + 63] = [01 10]
− 10 −5 +

As, the product of two matrices is identity matrix, so they are inverse to each other.

QN.3) (2077, New) If = [−41 12], find the value of A2 − 5A + 6I, where I be the identity

matrix of order 2×2.

Solution: Now, 2 − 5 + 6

= [−41 21] [−41 12] − 5 [−41 12] + 6 [01 10]

= [−164 − 2 −82++21] − [−205 150] + [06 60]
− 1

= [−145 −101] − [−145 −101]

= [00 00]

QN.4) (2077, Regular) Verify that: (AB)C = A(BC)

Where, = [13 42] , = [21 −03] , = [01 −11],

Solution: Now, = [31 42] [12 −03] = [31 + 4 00−−162] = [151 −−162]
+ 8

and = [12 −03] [10 −11] = [12 + 0 −1 + 03] = [21 −−15]
− 0 −2 −

LHS: (AB)C = [151 −−162] [10 −11] = [151−+00 −5 − 162] = [151 −−1213]
−11 −

RHS: A(BC) = [31 42] [21 −−51] = [31 + 4 −1 − 1200] = [151 −−1213]
+ 8 −3 −

∴ LHS = RHS

Hence proved

QN.5) (2076) Find the product AB of the following two matrices

278

=SSQA oo N=ll[ uu42.=tt6[ii++21)oo[nn(13512:: 130++ 7 6803 4 43])==][G84−[a[4i12n−++v14de139n13B++82A=]823443==][]212[[[15[6221−1244+201143−03+431−1]55363582=0]+563]+][a1−921n2401d]++=++B21[=404011]031[=15 −111[247213300]++,32s063012h]]06o]=w that AB ≠ BA.
[3149 −123]

∴ AB ≠ BA ==|[3553 2772|]=, th2e1n−fi1n0d=A−111

QN.7) (2074) If A
Solution: Now, | |

A11 = cofactor of 3 = 7, A12 = cofactor of 2 = -5

A21 = cofactor of 5 = -2, A22 = cofactor of 7 = 3
∴AAddAjjo−(1iAn=)t o=|fA1A[|−A=7d5j[(AAA11−)2132=]AA1122121[]−7=5[−7−532]−=32[]−75//1111
−2/11 ]
3/11
QN.8) (2072) Find the adjoint of matrix given below: = [13 −25]
Solution: Now, A11 = cofactor of 1 = -5, A12 = cofactor of 2 = -3

A21 = cofactor of 3 = -2, A22 = cofactor of -5 = 1
∴ Adjoint of A = [AA1112 wAAh1622e12|]re== 1[−−2=35−[211 −1261]]
= 11
QN. 9) (2072) Find A-1,
Solution: Now, | | = |21
A11 = cofactor of 2 = 6, A12 = cofactor of 1 = -1

A21 = cofactor of 1 = -1, A22 = cofactor of 6 = 2
= [AA1121 =AA1122121][−=61[−61−21]−2=1][−61//1111
Adjoint of A ( )

∴ −1 = 1 −1/11 ]
| | 2/11
=−[525 + 3060]] =−[4145] −21]
QN. 10) (2077, Back) Find (AB)T, where A −2 + and B
Soluti(oAnB:)ATB==[−[25205 301]41[]14 −21] = [280++30 = [1210

Descriptive Answer Question

279

QN. 11) (2077, Old) Solve the following linear equation by using determinant or matrix

method: 2x + 2y + 2z = 6, 4x + 2y + 3z = 10, x + 3y + 4z = 7

Solution: writing the given equation in matrix form we have
2 2 2 6
[4 2 3] [ ] = [10]
1 3 4 7
Which can be written as AX = B…………….. (i)
2 2 2 6
, where = [4 2 3] , = [10] , = [ ]
21 23 24 7
Now, | | = |4 2 3| = 2(8 − 9) − 2(16 − 3) + 2(12 − 2) = −2 − 26 + 20 = −8
134
A11= cofactor of 2 = 8 – 9 = -1, A12 = cofactor of 2 = -(16 – 3) = -13
A13 = cofactor of 2 = 12 – 2 = 10, A21= cofactor 4 = -(8 – 6) = -2
A22 = cofactor of 2 = 8 – 2 = 6, A23 = cofactor of 3 = - (6 – 2) = -4
A31 = cofactor of 1 = 6 – 4 = 2, A32 = cofactor of 3 = -(6 – 8) = 2
A33 = cofactor of 4 = 4 – 8 = -4
A11 A21 A31 −1 −2 2
Adj(A) = [A12 A22 A32] = [−13 6 2]
AAd1j3(A)A=23−18A[3−−311013 10 −4 −4
−2 2
A−1 = 1 6 2]
|A| −4 −4

From (i) X = A-1B −6 − 20 + 14 3/2
− 18=[−32−110,13 −2 2 6 [−78 + 60 + 14] −12 [1/2 ]
= 6 2 ] [10] = − 1 = − 1 [ −4 ] =
∴ =−124, −4 7 8 60 − 40 − 28 8 1
= −8

1

QN. 12) (2076) Prove that the two matrix ‘A’ and ‘B’ are inverse of each other, where
= =[31[31 21]21] [ − 1 1 =−32[−]1=1 [13−3−−2]12
−6 + 36] = [01 01]
−2 +
As, the product of two matrices is identity matrix, so they are inverse to each other.

QN.13) (2075) Evaluate A2 – 4A – 5I43w] [h21ere34[]12− 443][21 34] − 5 [10 01]
Solution: Now, 2 − 4 − 5 = [21

= [12 + 6 3 + 1162] − [48 1162] − [05 05]
+ 8 6 +

= [170 2125] − [98 2112]

= [−22 31]

280

QN. 14) (2074) An electronic company has obtained orders of 10 laptops of type A, 9 laptops

of type B, and 6 laptops of type C. The following matrix shows the amount of raw materials

needed for each types of laptops.

Laptops Raw Materials
R1 R2 R3

A 40 20 10

B 50 15 15

C 30 12 12

If the cost of each unit of R1, R2, and R3 are Rs.5000, Rs.6000, and Rs. 4000 respectively, find

by using matrix algebra.

a) The cost of the materials for laptop of each type.

b) The cost of the materials for all laptops.

Solution: Let the matrix of raw materials needed for each types of laptop be denoted

by X, cost of each type of raw material be denoted by Y, and number order of each

type laptop by Z then
40 20 10 5000
= [50 15 15] = [6000] and = [10 9 6]
30 12 12 4000

40 20 10 5000
Now, = [50 15 15] [6000]

30 12 12 4000
200000 + 120000 + 40000
= [ 250000 + 90000 + 60000 ]
150000 + 72000 + 48000
360000
= [400000]
270000

a) The cost of the R1, R2, and R3, is Rs.360,000, Rs. 400,000, and Rs. 270,000 respectively.

360000
Again, ( ) = [400000] [10 9 6] = [3600000 + 3600000 + 1620000] = [8820000

270000

b) The cost of materials for all laptops is Rs. 8,820,000.

QN. 15) (2074, Old) Solve the following equations by using matrix method
x + 2y + 3z = 20
3x + 6y + 2z = 32
x+y+z=9
Solution:
Now, writing the given equations in matrix form, we have

281

1 2 3 20
[3 6 2] [ ] = [32]
1 1 1 9

which can be written as AX = B

123 20 x
where A = [3 6 2] , B = [32] and X = [y]
1 1 1 −1 9 z
have = … … …… … . ( )
Again, we

Now,
123

| | = |3 6 2| = 1(6 − 2) − 2(3 − 2) + 3(3 − 6) = 4 − 2 − 9 = −7
111

Cofactor of 1 (( 1121))==4- 1
Cofactor of 2
Cofactor of 2 ( 13) = - 3

Cofactor of 263((( 222321)))===11-2
Cofactor of
Cofactor of

Cofactor of 111((( 333321)))===07-14
Cofactor of
Cofactor of

A11 A21 A31 4 1 −14
Adj(A) = [A12 A22 A32] = [−1 −2 7 ]
A13 A23 A33 −3 1 0

∴ −1 = 1 . ( ) = 1 4 1 −14
Form | | −7 [−1 −2 7]
−3 1 0
(i) = −1
−−11772[[−−−224131814] 1 −14 20
[ ] = −2 7 ] [32]
= 1 09
[ ]


[ ] = [−3]
−4

∴ x = 2, y = −3, z = −4 2 −1
5 4 1]
2 3
QN. 16) (2073, Old) From the given matrix = [2
1

282

Prove that AA-1= A-1A = I.

Solution: Now,
5 2 −1

| | = |2 4 1 | = 5(12 − 2) − 2(6 − 1) − 1(4 − 4) = 50 − 10 − 0 = 40
12 3

Cofactor of -251((( 11 1213)))===1-24(6–– 2 = 10
Cofactor of – 1) = -5
Cofactor of 4 = 0

Cofactor of 2 (( 2212)) = -(6 + 2) = - 8
Cofactor of 4 = 15 + 1 = 16
Cofactor of 1( 23) = - (10 – 2) = - 8

Cofactor of 1 ((( 333213))) = 2+4=6 7
Cofactor of 2 = - (5 + 2) = -
Cofactor of 3 = 20 – 4 = 16

A11 A21 A31 10 − 8 6
Adj(A) = [A12 A22 A32] = [−5 16 −7]
A13 A23 A33 0 − 8 16

∴ −1 = 1 . 422 (−13 1)]=41041[0−1[0−015005 −8 6
| | 16 −7]
−−88 616
5 16 −7]
−1 = [2 −8 16

1

= 1 50 − 10 + 0 − 40 + 32 + 8 30 − 14 − 16
60 [20 − 20 + 0 −16 + 64 − 8 12 − 28 + 16]
10 − 10 + 0 − 8 + 32 − 24 6 − 14 + 48

= 1 40 0 0
60 [0 40 0]
0 40
0

100
= [0 1 0]

001

−1 = 1 10 −8 65 2 −1
40 [−5 16 −7] [2 4 1]
−8 16 1 2 3
0

= 1 50 − 16 + 6 20 − 32 + 12 −10 − 8 + 18
40 [−25 + 32 − 7 −10 + 64 − 14 5 + 16 − 21 ]
0 − 8 + 48
0 − 16 + 16 0 − 32 + 32

= 1 1 0 0
40 [1 40 0]
0 0 40

283

100
= [0 1 0]

001
∴ AA-1 = A-1A= I

QN. 17) (2073) Solve the following equations by using elementary row operation method
2x + 5y +3z = 9
3x + y + 2z = 3
x + 2y - z = 6
Solution:
The augmented matrix of given system of equation is

1 5 3 : 9 , 1 → 1 1
∼ [3 2 2 : 23] 2
1 2
1 2 −1 : 6

1 5 3 : 9
2 2 2
13 5 21
∼0 − 2 − 2 : − 2 , 2 → 2 − 3 1, 3 → 3 − 1

[0 − 1 − 5 : 3 ]
2 2 2

1 5 3 : 9
2 2 2
5 21 2
∼0 1 13 : 13 , 2 → − 13 2

[0 − 1 − 5 : 3 ]
2 2 2

1 5 3 : 9
2 2 2
5 21 1
∼0 1 13 : 13 , 3 → 3 + 2 2

[0 0 − 30 : 30
13 13]

1 5 3 : 9 13
2 2 2 30
∼0 5 21 , 3 → − 3
[0 1 13 : −131]
0 1 :

284

1 5 0 :6 ], 2 → 2 − 5 3, 1 → 1 − 3 3
∼ [0 2 0 :2 13 2
1
0 0 1 : −1

1 0 0 :1 1 → 1 − 5 2
∼ [0 1 0 :2 ], 2
0 1 : −1
0

∴ = 1, = 2, = −1

QN. 18) (2072) A manufacturing company Ltd produces two types of drugs D1 and D2
with the help of three chemicals C1, C2, and C3. The quantity of chemicals requirements
per kg of each D1 and D2 are given below in suitable units.

D1 D2
C1 10 12
C2 16 15
C3 8 16
The price of chemicals in three different markets M1, M1 and M3 are as follows

C1 C2 C3
M1 20 19 16
M2 8 9 7
M3 6 7 8

Find the price per kg of each drug in each market by using matrix algebra.

Solution: Let A denote the matrix of quantity of chemicals requirement per kg of each type of

drug, and B denote the matrix of chemicals in three different markets M1, M1 and M3.

10 12 20 19 16
= [16 15] = [ 8 9 7 ]
8 16 678

20 19 16 10 12
Now, = [ 8 9 7 ] [16 15]

6 7 8 8 16

200 + 304 + 128 240 + 285 + 256
= [ 80 + 144 + 56 96 + 135 + 112 ]

60 + 112 + 64 72 + 105 + 128

632 781
= [280 343]

236 305

The price per kg of drug D1 in three different markets M1, M1 and M3 are respectively Rs.632,

Rs.280, Rs.236, and the price per kg of drug D2 in three different markets M1, M1 and M3 are

respectively Rs.781, Rs.343, Rs.305.

285

QN. 18) (2072) Solve by using matrix method

6x + 8y = 30

9x + 5y = 42

Solution: Now, writing the given equations in matrix form we have
[96 85] [ ] = [3420]

Which can be written as AX = B …………….. (i)
Where, = [69 58] , = [ ] , = [4320]
Now, | | = |96 85| = 30 − 72 = −42

Since, |A|≠ 0, so the system has unique solution.

A11 = cofactor of 6 = 5, A12 = cofactor of 8 = -9

A21 = cofactor of 9 = -8, A22 = cofactor of 5 = 6
= [AA1121 =AA−221214]2=[−5[9−59 −68]
Adjoint of A ( ) −68]

∴ −1 = 1
| |

From eq (i), we have X = A-1B

⇒ = 1 [−59 −68] [4320]
−42

⇒ = 1 [−125700−+225562]
−42

⇒ [ ] = 1 [−−168]
−42

∴ = 1 , = 3
7 7

QN. 19) (2072, Old) If = [13 12], then show that A2 – 2A – 5I = 0, where 0 and I are the 2 × 2

null matrix and I an identity matrix respectively.

Solution: Now, A2 − 4A − 5I = [13 12] [13 12] − 2 [31 21] − 5 [10 10]

= [13 + 6 2 + 21] − [62 42] − [05 50]
+ 3 6 +

= [67 47] − [76 70]

= [00 00]

= 0 Hence proved.

286

QN. 20) (2072, Old) If = [21 11], show that AA-1 = I.
Solution: Now, | | = |12 11| = 2 − 1 = 1A11 = cofactor of 2 = 1, A12 = cofactor of 1 = -1

A21 = cofactor of 1 = - 1, A22 = cofactor of 1 = 2
= [AA1121 AA1122[21−]11= [−11 −21]
Adjoint of A Adj(A) −21] = [−11

∴ A−1 = 1 = −21]
|A|

Again, −1 = [12 11] [−11 −21] = [21 − 1 −2 + 22] = [01 10]
− 1 −1 +

QN. 21) (2072, Old) A biscuits manufacturing company has two plants located at two different

places. Each plant produces three different types of biscuits Tiger, Glucose, and Salt. Matrix A

shows the number of packets of each types of biscuits produced per day and matrix B shows their

cost and selling price.

Tiger Glucose Salt SP CP
Plant I 300 200 100 10 13
Plant II 200 300 200 11 12
12 14

If the plant I and plant II operate for 20 and 25 days respectively in a month. Using matrix

algebra, find total

i) Production of biscuits from each plant per day.

ii) Production of each type of biscuits in the month.

iii) Profit in the month.

Solution: Here given

= [230000 200 120000] , = 10 13 25]
300 [11 12] = [20
12 14

i) Production of biscuits from plant plat I is 600 packets per day and from plat II is

700 packets per day.

ii) = [20 25] [320000 200 210000]
300

= [6000 + 5000 4000 + 7500 2000 + 5000]

= [11000 11500 7000]

287

Therefore, Manufacturing company produce 11000 packets of Tiger biscuit, 11500

packets of Glucose biscuit, and 7000 packets of Salt biscuit.

10 13
iii) ( ) = [11000 11500 7000] [11 12]

12 14
= [110000 + 126500 + 84000 143000 + 138000 + 98000]
= [320500 379000]
Finally, Profit = SP – CP = Rs.(320500 – 379000) = Rs. 58,500

QN. 22) (2071) The three commodities C1, C2, and C3 are purchased and sold by the three
persons A, B, and C. Mr. A purchase 4 units of C3 and sells 3 units of C1 and 5 units of C3. Mr.
B purchase 3 units of C1, and sells 2 units of C2 and 1 unit of C3. Mr. C purchase 1 unit of C2
and sells 4 units of C1 and 6 units of C3. In the purchase and sales process Mr. A loss of
Rs.1000, Mr. B earns no profit and Mr. C earns Rs. 40,000. Find the prices of per unit of these

commodities by using matrix method.

Solution: Solution:

Represent the given information in the table:
C1 C2 C3 Profit/loss

A 3 5 -4 -1000

B -3 2 1 0

C4 -1 6 40000

Let the prices of three commodities C1, C2, and C3 be x, y, and z respectively.
We can write the above information as follows
3x + 5y – 4z = -1000
-3x + 2y + z = 0
4x – y + 6z = 40000
Now, writing the given equations in matrix form, we have

3 5 −4 −1000
[−3 2 1 ] [ ] = [ 0 ]

4 −1 6 40000

which can be written as AX = B −1000 x
3 5 −4
where A = [−3 2 1 ] , B = [ 0 ] and X = [y]
4 −1 6 40000 z

Again, we have X = A−1B … … … … … . (i)
Now,

288

3 5 −4
| | = |−3 2 1 | = 3(12 + 1) − 5(−18 − 4) − 4(3 − 8) = 39 + 110 + 20 = 169

4 −1 6
Since, |A| ≠ 0, the system has unique solution.

Cofactor of 3 ( 11)= 12 +1 = 13
Cofactor of 5 ( 12) = - (-18 – 4) = 22
Cofactor of -4 ( 13) = 3 – 8 = -5

Cofactor of -3 ( 21) = -(30 + 4) = 34
Cofactor of 2 ( 22) = 18 + 16 = 34
Cofactor of 1 ( 23) = -(-3 – 20) = 23

Cofactor of 4 ( 31) = 5 + 8 = 13
Cofactor of -1 ( 32) = -(3 – 12) = 9
Cofactor of 6 ( 33) = 6 + 15 = 21

A11 A21 A31 13 34 13
Adj(A) = [A12 A22 A32] = [22 34 9 ]
A13 A23 A33 −5 23 21
1 1 13 34 13
∴ A−1 = |A| . Adj(A) = [22 34 9]
Form 169 −5 23 21

(i) = −1 13 −1000
[ ] 13 34 9 ][ 0 ]
= 1 [ 22 34 21 40000
169 −5 23

= 1161136990[0[−−835051204303257800000000000000000+]++0 0 + 520000
[ ] = 0 + 360000]
+ 840000
[ ]

[ ] = [2000]
5000
Therefore, the prices of three commodities C1, C2, and C3 is Rs. 3000, Rs.2000 and Rs.5000

respectively.

QN. 23) (2071, Old) Diary corporation is planning to produce 1200 kgs of baby food mixing 3
nutrients A, B, and C. The three nutrient components per kg cost Rs.12, Rs.16 and Rs.14
respectively. The final output of 1 kg of baby food pack must contains nutrient C to be twice that
of nutrient A. The budget available to buy component nutrients are the cooperation is Rs.18480.
Find amount of each nutrient which should mixed the final output by using matrix method.
Solution: Let amount the nutrient A contains x unit and nutrient contains y units, then
amount of nutrient C contains 2x units.
Now, we can write the above information as follows
x + y + 2x = 1200

289

or 3x + y = 1200 ………… (i)
12x + 16y + 28x = 18480

or 40x + 16y = 18480
5x + 2y = 2310 ………… (ii)
W[53riti21n]g[t h ]e=giv[21e23n10e00q]uations in matrix form we have

Which can be written as AX = B …………….. (i)
Where, = [53 21] , = [ ] , = [12231000]
Now, | | = |35 21| = 6 − 5 = 1

Since, |A|≠ 0, so the system has unique solution.

A11 = cofactor of 3 = 2, A12 = cofactor of 1 = -5

A21 = cofactor of 5 = - 1, A22 = cofactor of 2 = 3
= [AA1121 =AA112221[−]2=5 [−25 −31]
Adjoint of A ( ) −31] = [−25

∴ −1 = 1 −31]
| |

From eq (i), we have X = A-1B
⇒ = [−25 −31] [12321000]
⇒ = [−26400000−+26391300]
⇒ [ ] = [99300]
∴ x = 90, y = 930

Hence, three commodities A, B, and C required 90kg, 930kg and 180kg respectively.

QN. 24) (2070) The three different chemicals are mixed in a specific proportion to produce 100

units of a product. The company has to has to meet a constraint in the production of the product

that chemical X should be one third of the difference between chemicals Y and Z used. The cost

of the chemicals X, Y, and Z are respectively Rs.50, Rs.20 and Rs.40 and the total cost of all

chemicals is Rs.2900. Determine each unit of chemical X, Y, and Z used by matrix method.

Solution: Let the three chemicals X, Y, and Z respectively used be x unit, y unit, and z unit.

Then above information can be expressed as follows

x + y + z = 100 ………….. (i)

= 1 ( − )
3

290

or 3x - y + z = 0 …………. (ii)
50x + 20y + 40z = 2900
5x + 2y + 4z = 290 ……….. (iii)
Now, writing the given equations in matrix form, we have

1 1 1 100
[3 −1 1] [ ] = [ 0 ]
5 2 4 290

which can be written as AX = B ……………….. (i) x
111 100
where A = [3 −1 1] , B = [ 0 ] and X = [y]
15 12 14 290 z

Now, | | = |3 −1 1| = 1(−4 − 2) − 1(12 − 5) + 1(6 + 5) = −6 − 7 + 11 = −2
524
Since, |A| ≠ 0, the system has unique solution.

Cofactor of 1 ( 11)= - 4 – 2 = - 6
Cofactor of 1 ( 12) = - (12 - 5) = -7
Cofactor of 1 ( 13) = 6 + 5 = 11

Cofactor of 3 ( 21) = -(4 - 2) = -2
Cofactor of -1 ( 22) = 4 – 5 = -1
Cofactor of 1 ( 23) = -(2 - 5) = 3

Cofactor of 5 ( 31) = 1 + = 2
Cofactor of 2 ( 32) = -(1 - 3) = 2
Cofactor of 4 ( 33) = -1 – 3 = -4

A11 A21 A31 −6 −2 2
Adj(A) = [A12 A22 A32] = [−7 −1 2 ]
A13 A23 A33 11 3 −4

∴ A−1 = 1 . Adj(A) = 1 −6 −2 2
|A| −2 [−7 −1 2]
11 3 −4

From (i), we have X = A−1B

x = 1 −6 −2 2 100
[y] −2 [−7 −1 2 ][ 0 ]
z 3 −4 290
11

x = 1 −600 − 0 + 580
[y] −2 [ −700 − 0 + 580 ]
z 1100 + 0 − 1160

x = 1 −20
[y] −2 [−120]
z
−60

291

x 10
[y] = [60]
z 30
Therefore, the three chemicals X, Y, and Z respectively used be 10 unit, 60 unit, and 30 unit.

292