This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Math 18.02 (Spring 2009): Lecture 4
Square systems. Equations of planes February 10 Reading Material: From Simmons: 18.4. From Course Notes M. Last time: Matrices. Inverse Matrices
Today: Square systems. Word problem. How many solutions? Equations of planes 2 Recall: Matrix Inversion Given a n x in matrix A we recall the sequence of steps you need to obtain its inverse, assuming of
course that |Al # 0: (111 E12 a13 mll 77112 M13 mil —m12 77113 mil —m21 m2.1
{121 £122 023 -* mm 77122 W23 -* —m21 77122 —m23 —' —m12 m22 "maz 031 £132 033 17131 W32 mas 1713.1 —m32 mas W13 —m32 mas
A M 0 CT
minors cofactors transpose
Then l
A-1 = —CT.
lAl Remark. Why does it work? 1
mﬁlumn - a1277112 + a13m13) ) \ Ijlflﬁlzimn "— 622mm + 612377113)
cofator cap. of |A| where ﬁrst new ofA is replaced with 2’” mm — 2 identical news!
Then clet = 0. o~\ AA‘1 = magi") : ( More precisely let’s consider for example the entry at place 11 in Ail—1. Since we require that
AA‘1 = In, as indicated in the picture, for this entry we must have 1
—[a11m11 “ a12m12 + a13m13l = 1-
lAl This is indeed correct since allmn — (112mm + alamlg is exactly |A| computed using the expansion
with respect to the ﬁrst row.
On the other hand for the element at position 21 in AA‘1 we must have 1
Whammu — a22m12 + c23m13] = 0- Now notice that (12177111 — aggmlg + (12371113 is the determinant, also obtained by expanding with
respect to the ﬁrst rwo, but of the matrix obtained from A by replacing the ﬁrst row by the second
one. This determinant is in fact zero since the matrix has two equal rows. 3 Word Problem Next time I want to bring to school some beagles, croissants and donuts. I need to find out how
many of each kind I need to buy if I have the following requests : I There are about 250 students in the class I I am assuming that the number of students who would like a croissant is the same as the
number of students who would like a donut I I cannot carrie more that 36.5 kilos I 1 beagle: 200g, 1 croissant = 100g and 1 donut=150g.
Let’s translate the problem into equations. We call a a: = # beagles o y = # croissant o z = # donuts We also recall that 35 kilos: 35,000g. Now everything is expressed in the same unit. We are read}r
for the equations: a: + y + z = 250
055 + y — z = 0 (3.1)
200:1: + 1003} + 150z = 36, 500 Question: What is a: y and 2? To solve the problem we write everything using matrices. So if we define 1 1 1
A = 0 1 —1
200 100 150 250
B = 0
36, 500 and L0 then (3.1) can be written using matrix multiplication %
AX = B.
Clearly if A'1 can be found then
AX = B 4:» A-IAX = A‘lB 4:} ex = A‘lB <=> X = A‘lB, and this will give us the vector X and hence the answer to our question. First we have to compute
|A|. Since there is a zero in the second row, we compute the determinant with respect to it: |Al=1(150 — 200) + 1(100 — 200) = —150 s 0. Using the procedure recalled at the beginning of the lecture one can easily compute A‘1 and obtain _1 250 —50 —2
‘1 = — ~200 —50 1
150 —200 100 1
Then
_1 250 —50 -2 250 70
X=ﬁ —200 —50 1 - U = 90
—200 100 1 36,500 90
.1: =70 beagles
y = 90 croissants
2 =90 donuts 4 How many solutions to a system? Consider the system x+2y =5
2m+4y =11 Obviously there are no solutions to this system! What happens to the formula we presented above using matrices? If we write
A X B (3 illiHﬁl then why X = AﬁlB isn’t a solution?
Because A‘l doesn’t exist. Why? Because ]A| = 0. In this case we say that A is a singular matrix.
Now let’s change the example above just a little: m+2y =5
2m+4y :10 Notice that the second equation is just a multiple by 2 of the ﬁrst one, hence it doesn’t bring any
new information. So there are many solutions X = (3:, y), where a: is linked to y by the ﬁrst equation (or equivalently, the second), that is any couple X = (5 — 23}, y), for any real number 1 , solves the
system. There are inﬁnitely many solutions. In matrix notation we can say that the equality A X s
5 (iii-(EH10) has inﬁnitely many solutions of type for all real numbers y. Finally consider m+2y =0
2x+4y 20 Notice that again the second equation is just a multiple by 2 of the ﬁrst one. So any couple
X = (—239, y) for any real number 1; solves the system. Again there are inﬁnitely many solutions.
In matrix notation we can say that the equality has inﬁnitely many solutions of type
for all real numbers 1;. Summary: Given any n X n matrix A, II If A‘1 exists ( 4:} |A| #— 0) then for any n x 1 matrix B the system AX = B has a unique
solution X = A‘lB. o If A‘1 does not exist ( 4:} |Al = 0) then depending on the n x 1 matrix B 3:5 0, the system
may have no solutions or inﬁnitely many. 0 If A'1 does not exist ( 4:} |A| = 0) then if the n X 1 matrix B = O, the system has always
inﬁnitely many solutions. The following table should help you in remembering this discussion: (4.1) .. Exercise 1. Consider the system 23: + y + z _.
4:1; + 32 = (4.2)
a: «3- cy = 0 For which ualue(s) of C does the system have a non—zen) solution? We write the system as AX = B,
where
2 1 1
A = 4 0 3
1 c 0 and B = 0. If A‘1 exists, then clearly
X = A_1BA_1O = 0, so the system would admit only the zero solution. So we need to understand for which value of c
A has |A| = 0. Since if we replace the ﬁrst column with the ﬁrst column minus two times the last
and the second column with the second column minus the last the determinant does not change we
replace A with ~ 0 0 1
A = —2 -3 3
1 c 0 and we compute |A| = |A| = —2c+ 3 = 3 — 2c. We now set
detA=3——2c=0 4:» c=3/2 Summary: - IfdetA # 0 then X=A'IB=A‘1(§)=(§) I If detA = 0, since B = 0, we will have 00 many solutions. Since detA = 0 <=> c = g, the
answer to our problem is c = % Exercise 2. ( To do at home) With 0 = :3: above ﬁnd all the inﬁnitely many solutions of the system. 5 Equations of Planes In this section we start the analytic description of geometric objects in 3D. In this case analytic
means that the points P = (33,31, 2) on a geometric object are uniquely determined through a precise
equation or formula involving the unknown variables :6, y and 2:. We consider the tvpical problem: Given a point P0 and a vector IV. Find the plane through Po,
perpendicular to N. Easy Case: d
P0 = (0, 0, 0) and N = (cub, 0). Then in this case the plane is described by all points P = (3:, y, 2) such that the correspondent vector
_, a
OP = (3,1 ,2) is perpendicular to the given vector N: ....._.,. ———> _. —>
N_LOP 4:) N-0P=O. Since
—> —)
N-OP=am+by+cz=0, it follows that the (analytic) equation for our plane is
are + by + cz = 0. Remark. What we just did is to give a geometric interpretation to the inﬁnitely many solutions
P = (any, 2:) to the equation as: + by + cz = 0. So in particular the solution(s) to a 3 x 3 systems,
like the (4.2) we studied aboue, represents the intersection of three planes. General case: _.
P0 = (mo,yn,zg) and N = (a,b,c)_ Then in this case the plane is described by all points P = (3;, 39,2) such that the vector $5 = —. (a- — 590,1; —- yo, z — 20) is perpendicular to the given vector N: NLWD <=> N-ﬁ=0. Since
_. —>
N-PoP=a(:L‘—a:g)+b(y—yo)+c(z—zo) =0: it follows that the (analytic) equation for our plane is now (1(1: — :30) + My — ya) + C(z — 20) = 0. Handy Case: The equation of the plane is Study Guide 1. Answer the following questions: I Using the last remark. made above, can you interpret geometrically (using planes) table (4.1) ?
Think about parallel planes etc. 0 Can you come up with a word problem that can be solved with, a. (linear) system like the one
we considered in (3.1)? ...

View
Full Document

- Fall '08
- Auroux