Exam 1
Question 1 (5 points). Solve the svstem of eguations in any way vou sev fit, showing work
as needed:
3ry + dira =)
J'|-—.I‘_)?—--.)
ANSWER: There are NUMEROUS solution methods that are all basically equivalent.,
nelieding forming the angment natrix with rows [3 11} and {1 = 1{5] Many people simply
did elimination or substitution however, which here would take the forin of Equation 1 +
PEquation 2 to get Try == 21 for ry = 3. Ouce vou have £y = 3. it's easy to plug into
egquation 2 1o get e = <2 lor somtion rp - 4 and e o =20 Note tha! angimented nns
WILL give the same solution. as will substitution or other potential climinations.
- - .- - 4 em - -
Solve:
3z, +42, =1
) —T3=2>5
From the second equation:
=23+ 5
Substitute into the first:
3(xz2+5)+4zy =1
3z + 15+ 425 =1
Tz +15=1
Txs = —14
Ty = —2
Now plug back in:
z;—(-2)=5
z1+2=5
T =9
Answer
=3, =ZTg=-2
or as a vector: 1
, Question 2 (X poiuts). Consider the tollowing systew of equations. Does it Lave no
solutions, one solution. or infinite solutions. Show work to justify vour answer
P,
Ly 'y ~ Ly = 2
Joey —ap 4 By =3
Say 4 drg 4wy =7
ANSWER: Again. technieally wany niethods are possibie, bt tiee inrention is ro use
angniented matrices and row operations as below:
13 -1 |2 13 -1 2 L3 - R
=13 @310 =10 6 [=3})=(0 10 6 |-3
3 05 7 0O =10 6 |-3 0 0 0 10
Where the second matrix is #2 — 311 and 73 — 581 and the third is 73 - 12 meaning
that the equation is BOTH consistent AND has a free variable in ry. indicating infinite
solutions.
For Question 2, the system is:
Z) + 32y — 23 =2
3z — 2y + 323
=3
5z) + 5z,
+ 23 =17
Use the augmented matrix:
Lrarvare]
o e -
|
o lw
sel
e
|
N e
Row reduce:
Ry
— 3R,
s — 5R)
gives:
U T |
0 -10 6 |
0 -10 6 |
Then:
Ry~ Ry
gives:
¥ @ F |
0 -10 6 |
0 0|
2 Answer
The system is consistent because there is no contradiction row like:
0=5
But there is a free variable, because there are 3 variables and only 2 pivot rows.
So the system has:
infinitely many solutions
Key test:
free variable + consistent system = infinite solutions.
Question 1 (5 points). Solve the svstem of eguations in any way vou sev fit, showing work
as needed:
3ry + dira =)
J'|-—.I‘_)?—--.)
ANSWER: There are NUMEROUS solution methods that are all basically equivalent.,
nelieding forming the angment natrix with rows [3 11} and {1 = 1{5] Many people simply
did elimination or substitution however, which here would take the forin of Equation 1 +
PEquation 2 to get Try == 21 for ry = 3. Ouce vou have £y = 3. it's easy to plug into
egquation 2 1o get e = <2 lor somtion rp - 4 and e o =20 Note tha! angimented nns
WILL give the same solution. as will substitution or other potential climinations.
- - .- - 4 em - -
Solve:
3z, +42, =1
) —T3=2>5
From the second equation:
=23+ 5
Substitute into the first:
3(xz2+5)+4zy =1
3z + 15+ 425 =1
Tz +15=1
Txs = —14
Ty = —2
Now plug back in:
z;—(-2)=5
z1+2=5
T =9
Answer
=3, =ZTg=-2
or as a vector: 1
, Question 2 (X poiuts). Consider the tollowing systew of equations. Does it Lave no
solutions, one solution. or infinite solutions. Show work to justify vour answer
P,
Ly 'y ~ Ly = 2
Joey —ap 4 By =3
Say 4 drg 4wy =7
ANSWER: Again. technieally wany niethods are possibie, bt tiee inrention is ro use
angniented matrices and row operations as below:
13 -1 |2 13 -1 2 L3 - R
=13 @310 =10 6 [=3})=(0 10 6 |-3
3 05 7 0O =10 6 |-3 0 0 0 10
Where the second matrix is #2 — 311 and 73 — 581 and the third is 73 - 12 meaning
that the equation is BOTH consistent AND has a free variable in ry. indicating infinite
solutions.
For Question 2, the system is:
Z) + 32y — 23 =2
3z — 2y + 323
=3
5z) + 5z,
+ 23 =17
Use the augmented matrix:
Lrarvare]
o e -
|
o lw
sel
e
|
N e
Row reduce:
Ry
— 3R,
s — 5R)
gives:
U T |
0 -10 6 |
0 -10 6 |
Then:
Ry~ Ry
gives:
¥ @ F |
0 -10 6 |
0 0|
2 Answer
The system is consistent because there is no contradiction row like:
0=5
But there is a free variable, because there are 3 variables and only 2 pivot rows.
So the system has:
infinitely many solutions
Key test:
free variable + consistent system = infinite solutions.
