Math 1090
Quiz 3 - Extra credit
February 25th, 2015
Answer the following questions in the space provided. The value of every
question is indicated at the beginning. This is due on Monday, March 9.
No late assignments will be collected. You must turn in this sheet. This
assignment is OPTIONAL and will count as extra credit for Quiz 3.
Name:
UID:
1. Consider the following system of equations.
2x + y + t = a
x+t+w =b
y+z+t=c
x + 2y + z + w = d
x+z+t+w =e
(a) (1 point) Write down the matrix of coefficients A for this system.
(b) (7 points) Find A−1 (augment A with the identity and apply row operations).
(c) (2 points) Take as a, b, c, d, e the last 5 digits of your UID and solve the system
above using A−1 . (For instance, if your UID is 00123456, then you should take
a = 2, b = 3, c = 4, d = 5, e = 6).
(a) The system consists of 5 equations in 5 variables x, y, z, t, w, so the matrix of coefficients will have order 5 × 5. The first column will contain the coefficients of variable
x, the second one will contain the coefficients of y and so forth. We thus obtain


2 1 0 1 0
 1 0 0 1 1 



0
1
1
1
0
A=


 1 2 1 0 1 
1 0 1 1 1
(b) In order to find the inverse matrix,

2 1 0
 1 0 0

 0 1 1

 1 2 1
1 0 1
we augment A with the identity

1 0 1 0 0 0 0
1 1 0 1 0 0 0 

1 0 0 0 1 0 0 

0 1 0 0 0 1 0 
1 1 0 0 0 0 1
and we apply row operations until the matrix on the left becomes the identity. The
matrix that is left on the right will then be A−1 . The coefficients that we seek to
modify in each step are marked in red.


2
1
0
1
1
1
0
1
2
0
0
0
1
1
1
1
1
1
0
1
0
1
0
1
1
⇒
⇒
⇒
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
[3] ↔ [3] + [2]
[4] ↔ [4] − 2[3]
[5] ↔ [5] + [4]
0
0
0
0
1


[2] ↔ 2[2] − [1]
⇒
⇒
[4] ↔ [4] − [2]
[5] ↔ [5] − [4]
2
0
0
0
0
1
−1
0
0
0
0
0
1
−1
1
2
0
0
0
0
1
−1
0
0
0
0
0
1
0
0



⇒
⇒

[4] ↔ [4] + [3]
[5] ↔ [5] + [4]
2
0
0
0
0


⇒
[5] ↔ [5] − 3[4]

⇒
⇒
⇒
⇒
[1] ↔ [1] + [4]
[2] ↔ [2] + [4]
[3] ↔ [3] + 2[4]
[4] ↔ −[4]
⇒
⇒
[1] ↔ [1] + [2]
[2] ↔ −[2]
0
2
2
1
−2
0
0
1
−1
−2
0
0
0
1
1
0
2
2
2
−6
1
−1
−1
−1
3
0
2
2
1
−5
0
0
1
−1
1
0
0
0
1
−2
0
2
2
1
1
−1
0
0
0
0
0
1
0
0
1
1
2
−1
0
0
0
0
0
1
1
0
0
0
− 12
0
0


2
0
0
0
0
1
3
1
3
− 23
5
6
1
3
4
3
− 23
− 16
2
0
0
0
0
1
−1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
−1/2
−2/3
−1/3
−1
2/3
5/6
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
1
0
0
0
−1/2
−1
1/3
−1
2/3
5/6

2
0
0
0
0
[1] ↔ 12 [1]
[2] ↔ −[2]


1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
1
−1
−1
−1
0
1
−1
−1
−1
− 12

1
1
1
−1
1
0
2
2
2
0
0
2
2
2
1

0
0
1
1
0
0
0
1
−2
0
1
1
2
−1
0

⇒
⇒
1
1
2
−1
0
0
2
2
−1
−1
0
0
1
0
0
[5] ↔ − 61 [5]
[2] ↔ [2] − 2[5]
[3] ↔ [3] − 2[5]
[4] ↔ [4] − 2[5]
1
1
2
−1
−3
0
0
1
0
0
1
−1
−1
0
0
0
2
2
0
0
1
−1
0
0
0

⇒
⇒
⇒
1
−1
0
0
0
1
1
2
−3
0

1
−1
1
2
−2
2
0
0
0
0


⇒
2
0
0
0
0

⇒
0
0
0
1
0
0
0
0
0
1
0
0
1
−1
− 16
5
6
0
0
0
1
0
− 23
− 23
1
3
1
3
1/2
0
0
0
−1/2
−1/2
1/3
−1
2/3
5/6
0
0
1
0
0
1/2
−2/3
1
−1/3
−1/6

0
0
0
1
−1



0
0
0
0
1





0

1
3
1
3
1
3
− 61


1/3
−1/3
0
−1/3
1/3
0
1/3
0
−1/3
1/3
0
2
0
−1
0

0
0
0
0
− 61
1
3
−2/3
−1/3
0
2/3
−1/6
−1
1/3
0
2/3
−1/6
0
0
0
0
1
0
0
0
0
1
1
−1
0
0
0
0
2
0
0
0
1/3
2/3
1
−1/3
−1/6
1
−2/3
1
−1/3
−1/6
−1/2
1/3
0
2/3
−1/6




0
1/3
0
−1/3
1/3

Since we have obtained the identity matrix on the left hand side, we conclude that
the inverse matrix A−1 is the matrix on the right hand side, namely


1/2 −1/2 −1/2
0
1/2
 0
1/3
1/3
1/3 −2/3 



0
−1
0
0
1
A−1 = 


 0
2/3
2/3 −1/3 −1/3 
−1/2 5/6 −1/6 1/3 −1/6
Page 2
0
0
0
0
1


(c) We can always write our system in matrix form like

  
x
a
 y   b 

  
  
A
 z = c 
 t   d 
w
e
Multiplying both sides by A−1 we obtain



x
 y 




−1 
−1
=A 
z
A
A


| {z } 
 t 

I
w
a
b
c
d
e






so that the solution to our system is


  
x
a
1/2 −1/2 −1/2
0
1/2
 y 
 b   0
1/3
1/3
1/3
−2/3


  
 z  = A−1  c  =  0
−1
0
0
1


  
 t 
 d   0
2/3
2/3 −1/3 −1/3
w
e
−1/2 5/6 −1/6 1/3 −1/6






a
b
c
d
e






and you just need to multiply these 2 matrices, where a, b, c, d, e are numbers that
depend on your UID.
Page 3