Linear System and Equations Assignment

Bushra Alrashdi
Assignment HW1
1. Solve the system using elimination
6x−3y=9 5x−3y=6
6x-3y=9 ..........(1)
5x-3y=6...........(2)
(1)×5-(2) × 6 gives
30x-15y=45..........(3)
30x-18y=36............(4)
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-3y=-9
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y=3
substituting the value of “y” in (3), we get,
30x-15(3)=45
30x=45+45=90
x=3
therefore, x=y=3
2. Determine which of the points (−2,−4), (4,−5), and (−1,−1)lie on both the lines −4x1−5x2
= 9 and 3x1−x2 = −2.
Here we use trial and error method to find the points which lie on both the given lines.
Condition for points to be on line is that, they should satisfy the equation of lines.
Let us check for (-1,-1)
Given: -4x1-5x2=9
-4(-1)-5(-1) = 4+5=9
Given: 3x1-x2=-2
3(-1)-(-1)=-2

Therefore (-1,-1) lie on both the given lines.
Row echelon form:
The leading (first) entry in each row must be equal to one.
The leading entry on each subsequent row must be on a new column to the right.
All rows where all entries are zero are below rows where NOT all entries are zero.
3. Determine if the linear system −x1 − 2x2 = −2, − 7x2 = 0 is in echelon form
C. The system is not in echelon form because not every equation has a leading variable.
4. A linear system with three equations and two variables must be inconsistent
False.
Condition for system solutions:
If a1/a2 = b1/b2 = c1/c2 then, infinite solutions
If a1/a2 = b1/b2 ≠ c1/c2 then, no solution
If a1/a2 ≠ b1/b2, then, unique solution
5. For the following linear systems enter the letter U if the system has a unique solution, the
letter N if it has no solution, and the letter I if it has infinitely many solutions.
3x + 4y = 1, 3x − 4y = 1
Since, a1/a2 ≠ b1/b2, the system has unique solution
3x + 4y = 1, 6x + 8y = 1
Since, a1/a2 = b1/b2 ≠ c1/c2, the system has no solution
3x + 4y = 1, 6x + 8y = 2
Since, a1/a2 = b1/b2 = c1/c2, the system has infinitely many solutions.
6. 12x − 15y = 15, 20x − 25y = k
For the above system of equations to be consistent, k must equal

For the above system of equations to be consistent, the condition is,
a1/a2 = b1/b2 = c1/c2
Therefore, we have,
12/20 = -15/-25 = 15/k
3/5 = 3/5 = 15/k
3/5 = 15/k
K=(15*5)/3 = 25
Therefore, k=25
7. Give a geometric description of the following systems of equations
1. −x − 3y = −4 6x + 8y = −8 ?
Two lines intersecting at a point.
2. 4x + 4y = −2 10x + 10y = −5 ?
Two lines that are same.
3. 4x + 4y = −2 10x + 10y = −8
Two parallel lines.
8. Solve the following system of equations. If there are no solutions, type ”No Solution” for
both x and y. If there are infinitely many solutions, type ”x” for x, and an expression in
terms of x for y. 1x+2y = 2, −1x−2y = −2
Since, for a1/a2 = b1/b2 = c1/c2, infinite solutions
From given, we have,
1/-1 = 2/-2 = 2/-2
-1/1 = -1/1 = -1/1
Therefore, the given system of equations, has infinite solutions
x=x
2y=2-x
9. Determine which of the points (5,−4,−4), (6,−2,−1), and (1,−3,−2) satisfy the linear
system
2x1 − 5x2 − 2x3 = 21, −2x1 + 8x2 + 3x3 = −32
Here we use trial and error method to find the points which satisfy given system.
Condition for points to satisfy the system is that, they should satisfy the equation of lines.
Let us check for (1,-3,-2)

Given: 2x1 − 5x2 − 2x3 = 21
2(1) – 5(-3) – 2(-2) = 21
Given: −2x1 + 8x2 + 3x3 = −32
−2(1) + 8(-3) + 3(-2) = −32
Therefore (1,-3,-2) satisfy linear system.
10. Determine if the linear system x1 − x2 − x3 = 0, 2x3 = −4, 2x2 + 2x3 = 13 is in echelon
form
The system is not in echelon form because not every equation has a leading variable.
11. Find the set of solutions for the linear system x1 + 2x2 + 6x3 = 1, − x2 + 2x3 = 8, 5x3 = 5
5x3 = 5  x3=1
− x2 + 2x3 = 8  -x2+2(1) =8  -x2=8-2=6  x2=-6
x1 + 2x2 + 6x3 = 1  x1+2(-6)+6(1) =1  x1=7
12. Find the set of solutions for the linear system −2x1 + 5x2 + 3x3 = 11, − 6x2 − 9x3 = −7
x3 = x3
− 6x2 − 9x3 = −7  x2 = - (-7+9x3) / 6
−2x1 + 5x2 + 3x3 = 11 −2x1 + 5 [- (-7+9x3) / 6 ] + 3x3 = 11  x1 = - (27x3+31) /
12
If x3=0, then,
x2 = 7/6
x1 = - 31/12
13. For the following system to be consistent, k ≠
-3x-6y-6z=7