Applications of Matrices and Determinants Exercise 1.2 Solutions

Class 12 Samacheer Kalvi students, here are the solutions to Exercise 1.2 in Chapter 1 Applications of Matrices and Determinants. (Business Maths). You can find links to the other exercises also below.

Text Book Solutions Exercise 1.2

1 . Solve the following equations by using Cramer’s rule.
(i) 2x + 3y = 7; 3x + 5y = 9

$\Delta = \begin{vmatrix} 2&3 \\ 3&5 \end{vmatrix}\\$
$=10-9 \\$
$=1 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 7&3 \\ 9&5 \end{vmatrix}\\$
$=35-27 \\$
$=8 \\$

$\Delta_{y} = \begin{vmatrix} 2&7 \\ 3&9 \end{vmatrix}\\$
$=18-21 \\$
$=-3\\$

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{8}{1} \\$
$=8 \\$

$y =\dfrac{\Delta_{y}}{\Delta}\\$
$=\dfrac{-3}{1} \\$
$=-3 \\$

Therefore x = 8; y = -3

1 (ii) 5x + 3y = 17; 3x + 7y = 31

$\Delta = \begin{vmatrix} 5&3 \\ 3&7 \end{vmatrix}\\$
$=35-9 \\$
$=26 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 17&3 \\ 31&7 \end{vmatrix}\\$
$=119-93 \\$
$=26 \\$

$\Delta_{y} = \begin{vmatrix} 5&17 \\ 3&31 \end{vmatrix}\\$
$=155-51\\$
$=104\\$

$x =\frac{\Delta_{x}}{\Delta} \\$
$=\frac{26}{26} \\$
$=1 \\$

$y =\frac{\Delta_{y}}{\Delta} \\$
$=\frac{104}{26} \\$
$=4 \\$

Therefore x = 1; y = 4

1. (iii) 2x + y − z = 3, x + y + z = 1, x − 2y − 3z = 4

$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & -3 \end{vmatrix}\\$
$= 2 (-3 + 2) - 1 (-3 - 1) - 1 (-2 - 1) \\$
$= 2 (-1) -1 (-4) -1 (-3) \\$
$= -2 + 4 + 3 \\$
$=5 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & -3 \end{vmatrix}\\$
$= 3 (-3 + 2) - 1 (-3 - 4) - 1 (-2 - 4) \\$
$= 3 (-1) -1 (-7) -1 ( 6) \\$
$= -3 + 7 + 6 \\$
=10

$\Delta_{y} = \begin{vmatrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{vmatrix}\\$
$= 2 (- 3 - 4) -3 (-3 - 1) -1 (4 - 1) \\$
$= 2 (-7) - 3 (-4) - 1 (3) \\$
$= -14 + 12 - 3 \\$
=-5

$\Delta_{z} = \begin{vmatrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{vmatrix}\\$
$= 2 (4 + 2) - 1 (4 - 1) + 3 (- 2 - 1) \\$
$= 2 (6) - 1(3) + 3 (-3) \\$
$= 12 - 3 - 9 \\$
=0

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{10}{5} \\$
=5

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-5}{5} \\$
=-1

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{0}{5} \\$
=0

Therefore x = 2; y = -1; z = 0

(iv) x + y + z = 6, 2x + 3y − z = 5, 6x − 2y − 3z = − 7

$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 2 & 3 & -1 \\ 6 & -2 & -3 \end{vmatrix}\\$
$= 1 (-9 - 2) -1 (-6 + 6) + 1 (-4 - 18)\\$
$= 1 (-11) - 1 (0) +1 (-22) \\$
$= -11-22 \\$
$=-33 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 6 & 1 & 1 \\ 5 & 3 & -1 \\ -7 & -2 & 3 \end{vmatrix}\\$
$= 6 (- 9 - 2) - 1 (- 15 - 7) + 1 (- 10 + 21)\\$
$= 6 (-11) - 1 (-22)+ 1 (11) \\$
$= -66 + 22 + 11 \\$
=-33

$\Delta_{y} = \begin{vmatrix} 1 & 6 & 1 \\ 2 & 5 & -1 \\ 6 & -7 & -3 \end{vmatrix}\\$
$= 1 (-15 - 7) - 6 (- 6 + 6) + 1 (- 14 - 30) \\$
$= 1 (-22) - 6 (0) + 1 (-44) \\$
$= -22-44 \\$
=-66

$\Delta_{z} = \begin{vmatrix} 1 & 1 & 6 \\ 2 & 3 & 5 \\ 6 & -2 & -7 \end{vmatrix}\\$
$= 1 (- 21 + 10) - 1 (- 14 - 30) + 6 (- 4 - 18)\\$
$= 1 (-11) - 1 (-44) + 6 (-22) \\$
$= -11 + 44 - 132 \\$
=-99

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-33}{-33} \\$
=1

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-66}{-33} \\$
=2

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{-99}{-33} \\$
=3

Therefore x = 1; y = 2; z = 3

1. (v) x + 4y + 3z = 2, 2x − 6y + 6z = −3, 5x − 2y + 3z = −5

$\Delta = \begin{vmatrix} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{vmatrix}\\$
$= 1 (-18 + 12) -4 (6 -30) + 3 (-4 + 30) \\$
$= 1 (-6) - 4 (-24) + 3 (26) \\$
$= -6 + 96 + 78 \\$
$=168 \neq 0\\[.25cm]$

$\Delta_{x} = \begin{vmatrix} 2 & 4 & 3 \\ -3 & -6 & 6 \\ -5 & -2 & 3 \end{vmatrix}\\$
$= 2 (-18 + 12) - 4 (-9 + 30) + 3 (6 - 30) \\$
$= 2 (-6) -4 (21) + 3 (-24) \\$
$= -12 - 84 - 72 \\$
=-168

$\Delta_{y} = \begin{vmatrix} 1 & 2 & 3 \\ 2 & -3 & 6 \\ 5 & -5 & 3 \end{vmatrix}\\$
$= 1 (-9 + 30) - 2 (6 - 30) + 3 (-10 + 15) \\$
$= 1(21) - 2 (-24) + 3 (5) \\$
$= 21 + 48 + 15 \\$
=84

$\Delta_{z} = \begin{vmatrix} 1 & 4 & 2 \\ 2 & -6 & -3 \\ 5 & -2 & -5 \end{vmatrix}\\$
$= 1 (30 - 6) - 4 (-10 +15) + 2 (-4 + 30) \\$
$= 1 (24) - 4 (5) + 2 (26) \\$
$= 24 - 20 + 52 \\$
=56

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-168}{168} \\$
=-1

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{84}{168} \\$
$=\dfrac{1}{2} \\$

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{56}{168} \\$
$=\dfrac{1}{3} \\$

Therefore $x = -1;$ $y =\frac{1}{2};$ $z =\frac{1}{3}$

2. A commodity was produced by using 3 units of labour and 2 units of capital, the total cost is Rs 62. If the commodity had been produced by using 4 units of labour and one unit of capital, the cost is Rs 56. What is the cost per unit of labour and capital? (Use determinant method).

Let x be the cost per unit of labour and y be the capital.
3x + 2y = 62
4x + y = 56

$\Delta = \begin{vmatrix} 3&2 \\ 4&1 \end{vmatrix}\\$
$=3-8 \\$
$=-5 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 62&2 \\ 56&1 \end{vmatrix}\\$
$=62-112$
=-50

$\Delta_{y} = \begin{vmatrix} 3&62 \\ 4&56 \end{vmatrix}\\$
$=168-248$
=-80

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-50}{-5} \\$
=10

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-80}{-5} \\$
=16

Therefore, the cost per unit of labour is Rs. 10 and cost per unit of capital is Rs. 16.

3. A total of Rs 8,600 was invested in two accounts. One account earned $=4 \frac{3}{4}%$
annual interest and the other earned $=6 \frac{1}{2}%$ annual interest. If the total interest for one year was Rs. 431.25, how much was invested in each account? (Use determinant method)

Let the two investments be x and y.

$x + y = 8600 \cdots (1) \\$
$=\dfrac{(4\frac{3}{4})}{100}x + \dfrac{(6\frac{1}{2})}{100}y = 431.25 \\$
$=\dfrac{(\frac{19}{4})}{100}x + \dfrac{(\frac{13}{2})}{100}y = 431.25 \\$
$=\dfrac{19}{4}x + \dfrac{13}{2} y = 43125$

Multiply by 4 on both sides

$19x + 26y = 172500 \cdots (2) \\$

$\Delta = \begin{vmatrix} 1&1 \\ 19&26 \end{vmatrix}\\$
$=26-19 \\$
$=7 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 8600&1 \\ 172500&26 \end{vmatrix}\\$
$=223600-172500 \\$
=51100

$\Delta_{y} = \begin{vmatrix} 1&8600 \\ 19&172500 \end{vmatrix}\\$
$=172500-163400 \\$
=9100

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{51100}{7} \\$
=7300

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{9100}{7} \\$
=1300

Therefore, the amount invested at $=4 \frac{3}{4} %$ annual interest is Rs. 7300 and and the amount invested at $=6 \frac{1}{2} %$ annual interest is Rs. 1300.

4. At marina two types of games viz., Horse riding and Quad Bikes riding are available on hourly rent. Keren and Bentia spent Rs 780 and Rs 560 during the month of May.

Name	Number of hours		Total amount spent in Rs.
	Horse Riding	Quad Bike Riding
Keren	3	4	780
Benita	2	3	560

Find the hourly charges for the two games (rides). (Use determinant method).

Let the hourly charges for horse riding be x and let the hourly charges for quad bike riding be y.

$3x + 4y = 780 \cdots (1) \\$
$2x + 3y = 560 \cdots (2)$

$\Delta = \begin{vmatrix} 3&4 \\ 2&3 \end{vmatrix}\\$
$=9-8 \\$
$=1 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 780&4 \\ 560&3 \end{vmatrix}\\$
$=2340-2240 \\$
=100

$\Delta_{y} = \begin{vmatrix} 3&780 \\ 2&560 \end{vmatrix}\\$
$=1680-1560 \\$
=120

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{100}{1} \\$
=100

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{120}{1} \\$
=120

Therefore, the hourly charges for horse riding is Rs. 100 and quad bike riding is Rs. 120.

5. In a market survey three commodities A, B and C were considered. In finding out the index number some fixed weights were assigned to the three varieties in each of the commodities. The table below provides the information regarding the consumption of three commodities according to the three varieties and also the total weight received by the commodity.

Commodity Variety	Variety			Total weight
	I	II	III
A	1	2	3	11
B	2	4	5	21
C	3	5	6	27

Find the weights assigned to the three varieties by using Cramer’s Rule.

Let x, y and z be are consumption of the three commodities A, B and C respectively.

$x + 2y + 3z = 11 \cdots (1) \\$
$2x + 4y + 5z = 21 \cdots (2) \\$
$3x + 5y + 6z = 27 \cdots (3) \\$

$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{vmatrix}\\$
$= 1 (24 - 25) - 2 (12 - 15) + 3 (10- 12)\\$
$= 1 ( 1) 2 ( 3) + 3 (- 2) \\$
$= -1 + 6 - 6 \\$
$=-1 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{vmatrix}\\$
$= 11 (24 - 25) -2 (126 - 135) + 3 (105 - 108) \\$
$= 11(-1) -2 (-9)+ 3 (-3) \\$
$= -11 + 18 - 9 \\$
=-2

$\Delta_{y} = \begin{vmatrix} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{vmatrix}\\$
$= 1 (126 - 135) -11 (12 - 15) + 3 (54 - 63) \\$
$= 1 (-9) -11 (-3) + 3 (-9) \\$
$= -9 + 33 - 27 \\$
=-3

$\Delta_{z} = \begin{vmatrix} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{vmatrix}\\$
$= 1 (108 - 105) - 2 (54 - 63) + 11 (10 - 12) \\$
$= 1(3) - 2 (-9) + 11 (-2) \\$
$= 3 + 18 - 22 \\$
=-1

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-2}{-1} \\$
=2

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-3}{-1} \\$
=3

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{-1}{-1} \\$
=1

Therefore x = 2; y = 3; z = 1

6. A total of Rs. 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6% if the total simple interest for one year was Rs. 380 and the amount invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule).

Let the amount invested at 2%, 3%, and 6% be x, y, and z respectively.

$x + y +z = 8500 \cdots (1) \\$
$= x\times\dfrac{2}{100} + y \times \dfrac{3}{100} + z\times \dfrac{6}{100} = 380\\$
$=\dfrac{2x}{100} + \dfrac{3y}{100} + \dfrac{6z}{100} =380 \\$
$2x + 3y + 6z = 38000 \cdots (2) \\$
$x + y = z \\$
$x + y - z = 0 \cdots (3)\\$

$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{vmatrix}\\$
$= 1 (-3 - 6) -1 (-2 - 6) + 1 (2 - 3) \\$
$= 1 (-9) -1 (-8) + 1 (-1) \\$
$= -9 + 8 - 1 \\$
$=-2 \neq 0\\$

$\Delta_{x} = \begin{vmatrix} 8500& 1 & 1 \\ 38000& 3 & 6 \\ 0& 1 & -1 \end{vmatrix} \\$
$= 8500 (-3 - 6) -1 (-38000 - 0) + 1 (38000 - 0) \\$
$= 8500 (-9) + 38000 + 38000 \\$
$= -76500 + 76000 \\$
=-500

$\Delta_{y} = \begin{vmatrix} 1 & 8500 & 1 \\ 2 & 38000 & 6 \\ 1 & 0 & -1 \end{vmatrix}\\$
$= 1(-38000 - 0) - 8500 (-2 - 6) + 1 (0 - 38000) \\$
$= -38000 + 68000-38000 \\$
$= 68000 - 76000 \\$
=-8000

$\Delta_{z} = \begin{vmatrix} 1 & 1 & 8500 \\ 2 & 3 & 38000 \\ 1 & 1 & 0 \end{vmatrix}\\$
$= (0 - 38000) -1 (0 - 38000) + 8500 (2 - 3)\\$
$= -38000 + 38000 - 8500 \\$
=-8500

$x =\dfrac{\Delta_{x}}{\Delta} \\$
$=\dfrac{-500}{-2} \\$
=250

$y =\dfrac{\Delta_{y}}{\Delta} \\$
$=\dfrac{-8000}{-2} \\$
=4000

$y =\dfrac{\Delta_{z}}{\Delta} \\$
$=\dfrac{-8500}{-2} \\$
=4250

Class 12 Applications of Matrices and Determinants Exercise 1.2 Solutions

Text Book Solutions Exercise 1.2

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