StudentAidUnit 3: Matrices and Determinants — Exercise 3 6
10th Class Mathematics · Unit 3: Matrices and Determinants
3.6.1.Solve by matrix inversion method, if possible. (i) 2x+5y=19, 4x-3y=-1 (ii) 3x+2y=7, 5x-y=16 (iii) x-2y=9, 2x+7y=-4 (iv) 3x+2y=2, x-2y=-2
(i)
A=begin{bmatrix}2&54&-3end{bmatrix}, |A|=-26, X=A-1B=begin{bmatrix}23end{bmatrix}
(ii)
A=begin{bmatrix}3&25&-1end{bmatrix}, |A|=-13, X=A-1B=begin{bmatrix}3-1end{bmatrix}
(iii)
A=begin{bmatrix}1&-22&7end{bmatrix}, |A|=11, X=A-1B=begin{bmatrix}5-2end{bmatrix}
(iv)
A=begin{bmatrix}3&21&-2end{bmatrix}, |A|=-8, X=A-1B=begin{bmatrix}01end{bmatrix}
(i) x=2,y=3 (ii) x=3,y=-1 (iii) x=5,y=-2 (iv) x=0,y=1
3.6.2.Use Cramer's rule to solve the following pair of linear equations, if possible. (i) x+4y=4, 2x-y=5 (ii) x+2y=7, 3x-2y=-3 (iii) 2x-5y=-6, 4x-3y=-12 (iv) 3x+2y=-1, 5x+6y=5
(i)
D=-9, Dx=-24, Dy=-3 ⇒ x=8/3, y=1/3
(ii)
D=-8, Dx=-8, Dy=-24 ⇒ x=1, y=3
(iii)
D=14, Dx=-42, Dy=0 ⇒ x=-3, y=0
(iv)
D=8, Dx=-16, Dy=20 ⇒ x=-2, y=5/2
(i) x=8/3,y=1/3 (ii) x=1,y=3 (iii) x=-3,y=0 (iv) x=-2,y=5/2
3.6.3.An electrical engineer wants to determine the current in two branches A and B of a simple electrical circuit. The system of the equations is: x+y=7, 2x-y=2, where x is the current in branch A and y is the current in branch B. Find x and y by using matrices.
Given
System
x+y=7,quad 2x-y=2
A=begin{bmatrix}1&12&-1end{bmatrix}, |A|=-3, A-1=begin{bmatrix}1/3&1/32/3&-1/3end{bmatrix}
Working
X=A-1B=begin{bmatrix}1/3&1/32/3&-1/3end{bmatrix}begin{bmatrix}72end{bmatrix}=begin{bmatrix}34end{bmatrix}
Result
Answer
x=3, y=4
3.6.4.Three forces act on a particle and must be in equilibrium i.e. F1+F2+F3=0, where F1=[[8],[x]], F2=[[-2],[-7]], F3=[[y],[-1]]. Find the value of x and y.
Given
Given
F1+F2+F3=0
Adding
begin{bmatrix}8-2+yx-7-1end{bmatrix}=begin{bmatrix}00end{bmatrix} ⇒ begin{bmatrix}6+yx-8end{bmatrix}=begin{bmatrix}00end{bmatrix}
Result
Answer
6+y=0 ⇒ y=-6;quad x-8=0 ⇒ x=8
3.6.5.Two support beams, A and B are holding up a combined load of 100 kN. Twice the load on beam A and three times the load on beam B equals 240 kN. Find the load of beam A and beam B by using matrices.
Given
Let loads
a+b=100,quad 2a+3b=240
A=begin{bmatrix}1&12&3end{bmatrix}, |A|=1, A-1=begin{bmatrix}3&-1-2&1end{bmatrix}
Working
X=A-1B=begin{bmatrix}3&-1-2&1end{bmatrix}begin{bmatrix}100240end{bmatrix}=begin{bmatrix}6040end{bmatrix}
Result
Answer
Load on beam A=60 kN, B=40 kN
3.6.6.In a 2D game world, two characters are moving along straight paths. One character moves along a line where the total of twice their horizontal position and vertical position is 5, while the other moves along a line where their horizontal position is one more than their vertical position. Find their point of intersection by using matrices.
Given
Equations
2x+y=5,quad x-y=1
A=begin{bmatrix}2&11&-1end{bmatrix}, |A|=-3, A-1=begin{bmatrix}1/3&1/31/3&-2/3end{bmatrix}
Working
X=A-1B=begin{bmatrix}1/3&1/31/3&-2/3end{bmatrix}begin{bmatrix}51end{bmatrix}=begin{bmatrix}21end{bmatrix}
Result
Answer
Point of intersection is (2,1)
3.6.7.Two years ago a man was 5 times as old as his son was. After 6 years he will be 3 times as old as his son. Find their present ages by using matrices.
Given
Let present ages
Man=x, Son=y
Formula
Two years ago
x-2=5(y-2) ⇒ x-5y=-8
After 6 years
x+6=3(y+6) ⇒ x-3y=12
A=begin{bmatrix}1&-51&-3end{bmatrix}, |A|=2, A-1=12begin{bmatrix}-3&5-1&1end{bmatrix}
Working
X=A-1B=12begin{bmatrix}-3&5-1&1end{bmatrix}begin{bmatrix}-812end{bmatrix}=begin{bmatrix}4210end{bmatrix}
Result
Answer
Man=42 years, Son=10 years
3.6.8.Two cyclists are 44 km apart and start out at the same time. If they go towards one another they meet in 2 hours, but if they go in the same direction the faster overtakes the slower in 7 1/2 hours. Find their speeds by using matrices.
Given
Let speeds
faster=x km/h, slower=y km/h
Formula
Towards each other
(x+y) × 2=44 ⇒ x+y=22
Same direction
(x-y) × 712=44 ⇒ x-y=44 × 215=8815
A=begin{bmatrix}1&11&-1end{bmatrix}, |A|=-2, A-1=begin{bmatrix}1/2&1/21/2&-1/2end{bmatrix}
Working
X=A-1B=begin{bmatrix}1/2&1/21/2&-1/2end{bmatrix}begin{bmatrix}2288/15end{bmatrix}=begin{bmatrix}209/15121/15end{bmatrix}
Result
Answer
Speeds are x=20915 km/h (faster), y=12115 km/h (slower)
3.6.9.The numerator of a fraction is 7 less than the denominator. If the numerator is increased by 3, the new fraction can be cancelled down to 3/4. Find the original fraction by using matrices.
Given
Let denominator
x;quad numerator=x-7
Formula
After increasing numerator by 3
x-7+3x=34 ⇒ x-4x=34
Cross multiply
4(x-4)=3x ⇒ 4x-16=3x ⇒ x=16
Numerator=16-7=9
Result
Answer
Original fraction is 916