Cramer's Rule to Solve Linear Equations SEE Optional Mathematics Solution Nepal
Exercise 3.4
Page 120
Page 120
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In the equation a1x + b1y = c1 a2x + b2y = c2 , according to Crammer’s rule,
a) Write D in determinant form.
D =
a1
a2
b1
b2
= a1b2 - a2b1
b) Write Dx in determinant form.
Dx =
c1
c2
b1
b2
= c1b2 - c2b1
c) Write Dy in determinant form.
Dy =
a1
a2
c1
c2
= a1c2 - a2c1
d) What is necessary condition for obtaining unique solution?
D ≠ 0 is necessary condition for obtaining unique solution.
- Solve the following system of linear equations using Cramer’s rule:
a)
2x - y = 3
y + 3x = 7
y + 3x = 7
Solution
2x - y = 3 ....(i)
3x + y = 7 ...(ii)
D =
2
3
-1
1
= 2×1 - 3×(-1) = 2 - (-3) = 2 + 3 = 5
D₁ =
3
7
-1
1
= 3×1 - 7×(-1) = 3 - (-7) = 3 + 7 = 10
D₂ =
2
3
3
7
= 2×7 - 3×3 = 14 - 9 = 5
x =
D₁
D =
10
5 = 2
y =
D₂
D =
5
5 = 1
D₁
D =
10
5 = 2
y =
D₂
D =
5
5 = 1
b)
y = 2x
x -
3
2y + 1 = 0
3
2y + 1 = 0
Solution
y = 2x
2x - y = 0 ....(i)
x -
3
2y + 1 = 0
Or, 2x - 3y + 2 = 0
2x - 3y = -2 ....(ii)
D =
2
2
-1
-3
= 2×(-3) - 2×(-1) = -6 - (-2) = -6 + 2 = -4
D₁ =
0
-2
-1
-3
= 0×(-3) - (-2)×(-1) = 0 - 2 = -2
D₂ =
2
2
0
-2
= 2×(-2) - 2×0 = -4 - 0 = -4
x =
D₁
D =
-2
-4 = 0.5
y =
D₂
D =
-4
-4 = 1
D₁
D =
-2
-4 = 0.5
y =
D₂
D =
-4
-4 = 1
c)
4
x +
5
y = 28
7
x +
3
y = 67
D =
4
7
5
3
= 4×3 - 7×5 = 12 - 35 = -23
D₁ =
28
67
5
3
= 28×3 - 67×5 = 84 - 335 = -251
D₂ =
4
7
28
67
= 4×67 - 7×28 = 268 - 196 = 72
1
x =
D₁
D
Or,x =
D
D₁ =
-23
-251 = 0.092
1
y =
D₂
D
Or,y =
D
D₂ =
-23
72 = -0.319
4
7
5
3
= 4×3 - 7×5 = 12 - 35 = -23
D₁ =
28
67
5
3
= 28×3 - 67×5 = 84 - 335 = -251
D₂ =
4
7
28
67
= 4×67 - 7×28 = 268 - 196 = 72
1
x =
D₁
D
Or,x =
D
D₁ =
-23
-251 = 0.092
1
y =
D₂
D
Or,y =
D
D₂ =
-23
72 = -0.319
d)
4( x – 1) + 5(y + 2) = 10
5( x – 1) – 3(y + 2) + 6 = 0
5( x – 1) – 3(y + 2) + 6 = 0
Solution
4(x-1) + 5(y+2) = 10
Or, 4x - 4 + 5y + 10 = 10
4x + 5y = 10 - 10 + 4
4x + 5y = 4 ....(i)
5(x-1) - 3(y+2) + 6 = 0
Or, 5x - 5 - 3y -6 + 6 = 0
5x - 3y = 5
5x - 3y = 5 ....(ii)
D =
4
5
5
-3
= 4×(-3) - 5×5 = -12 - 25 = -37
D₁ =
4
5
5
-3
= 4×(-3) - 5×5 = -12 - 25 = -37
D₂ =
4
5
4
5
= 4×5 - 5×4 = 20 - 20 = 0
x =
D₁
D =
-37
-37 = 1
y =
D₂
D =
0
-37 = 0
D₁
D =
-37
-37 = 1
y =
D₂
D =
0
-37 = 0
e)
3xy – 10y = 6x
5xy + 3x = 21y
5xy + 3x = 21y
Solution
3xy - 10y = 6x
Or, 6x + 10y = 3xy
Dividing both sides by xy
Or,
6x
xy +
10y
xy =
3xy
xy
6
y +
10
x = 3
10
x +
6
y = 3 ....(i)
5xy + 3x = 21y
Or, 3x - 21y = -5xy
Dividing both sides by xy
Or,
3x
xy -
21y
xy =
-5xy
xy +
3
y -
21
x = -5
-21
x +
3
y = -5 ....(ii)
D =
10
-21
6
3
= 10×3 - (-21)×6 = 30 - (-126) = 30 + 126 = 156
D₁ =
3
-5
6
3
= 3×3 - (-5)×6 = 9 - (-30) = 9 + 30 = 39
D₂ =
10
-21
3
-5
= 10×(-5) - (-21)×3 = -50 - (-63) = -50 + 63 = 13
1
x =
D₁
D
Or,x =
D
D₁ =
156
39 = 4
1
y =
D₂
D
Or,y =
D
D₂ =
156
13 = 12
f)
3y + 4x = 2xy and
18y – 4x = 5xy
18y – 4x = 5xy
Solution
3y + 4x = 2xy
Dividing both sides by xy
Or,
3y
xy +
4x
xy =
2xy
xy +
3
x +
4
y = 2 ....(i)
18y - 4x = 5xy
Dividing both sides by xy
Or,
18y
xy -
4x
xy =
5xy
xy +
18
x -
4
y = 5 .....(ii)
D =
3
18
4
-4
= 3×(-4) - 18×4 = -12 - 72 = -84
D₁ =
2
5
4
-4
= 2×(-4) - 5×4 = -8 - 20 = -28
D₂ =
3
18
2
5
= 3×5 - 18×2 = 15 - 36 = -21
1
x =
D₁
D
Or,x =
D
D₁ =
-84
-28 = 3
1
y =
D₂
D
Or,y =
D
D₂ =
-84
-21 = 4
- Ask the price of any five daily uses goods. Make two different system of equation a and b in terms of x and y. Solve these equations by Cramer’s rule and present your findings in the classroom.
Solution
1 L milks costs Rs.80
1 L curd costs Rs.130
My mom bought 0.5L milk and 1L curd and paid Rs.170 on last Sunday
This Tuesday she paid Rs.550 for 2L milk and 3L curd. She does not know the rate of milk and curd. She told me to find the rate. Let x represent rate of milk and y represents rate of curd.
0.5x + y = 170
2x + 3y = 550
D =
0.5
2
1
3
= 0.5×3 - 2×1 = 1.5 - 2 = -0.5
D₁ =
170
550
1
3
= 170×3 - 550×1 = 510 - 550 = -40
D₂ =
0.5
2
170
550
= 0.5×550 - 2×170 = 275 - 340 = -65
x =
D₁
D =
-40
-0.5 = 80
y =
D₂
D =
-65
-0.5 = 130
D₁
D =
-40
-0.5 = 80
y =
D₂
D =
-65
-0.5 = 130
Rate of milk is found to be Rs.80 for 1L and rate of curd is found to be Rs.130 which is same as the market price.What we find here that cramers rule can be used to solve such daily life problems.
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