Filetype PDF | Posted on 27 Jan 2023 | 2 years ago
Authentication
digital resources
176x Filetype PDF File size 0.32 MB Source: www.math.hkust.edu.hk
Matrices (Ch.4)
(i) System of linear equations in 2 variables (L.5, Ch4.1)
◮ Find solutions by graphing
MATH1003 Review: Part 2. Matrices ◮ Supply and demand curve
(ii) Basic ideas about Matrices (L.6, Ch4.2)
◮ To know a matrix
Maosheng Xiong
Department of Mathematics, HKUST
◮ Row operation
R ↔R, kR →R, R +kR →R
i j i i i j i
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Matrices (Ch.4) Matrices (Ch.4)
(iii) Gauss-Jordan Elimination (Method 1 to solve systems of (v) Inverse Matrix (L.10, Ch4.5)
linear eqns.) ◮ Identity matrix I
◮ Find the corresponding augmented matrix (L.7, Ch4.3) IM =MI =M
◮ Using row operation to get the reduced form (L.7, Ch4.3) −1
◮ Unique solution / no solution / infinite number of solutions ◮ Given a square matrix M, its inverse matrix M satisfies
(how many degrees of freedom) (L.7, Ch4.3) MM−1=M−1M=I
◮ Application: variables and restrictions, positive numbers may
be required for some real world problems (L.8, Ch4.3) ◮ Find the inverse matrix M−1:
(iv) Matrix Operation (L.9, Ch4.4) ⊲Oforder 2: M = a b, then M−1 = 1 d −b
c d ad−bc −c a
⊲Oforder 3: (M|I)Rowoperation(I|M−1)
−→
⊲Notall matrices are invertible
(vi) Matrix Equation (L.11, Ch4.6)
◮ Solver to system of linear equations (method 2 to solve
systems of linear eqns): if A is invertible,
−1
AX =B ⇒ X=A B.
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Matrices (Ch.4) Problems and Solutions
(vii) Leontief input-output analysis (L.12, Ch.4.7)
◮ Technology matrix M for n-industry is constructed:
Example
Given
A= 1 3 , B= 0 1 −5 ,
2 0 −2 −1 2
◮ The final demand matrix D and the output matrix X are n ×1 C =AB. Find the second column of matrix C.
matrices.
◮ The matrix equation for the model is
X = M X + D .
|{z} |{z} |{z}
output technology matrix external demand
◮ −1
Therefore, X = (I −M) D if I − M is invertible.
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Problems and Solutions Problems and Solutions
Solution Example
A is of size 2 × 2 and B is of size 2 × 3, so C is of size 2 × 3. For Given a system of linear equations:
its second column, we only need to calculate c and c .
12 22
◮ For c , the 1st row of A and the 2nd column of B are needed: x −x +2x =1
12 1 2 3
x +x +4x =5
1 1 2 3
c = 1 3 =1·1+3·(−1)=−2.
12 2x +hx =k
−1 1 3
◮ For c , the 2nd row of A and the 2nd column of B are
22 (a) Write down the corresponding augmented matrix (A|b) from
needed: Ax =b.
1 (b) For what value of h, the system can not have a unique solution
c22 = 2 0 −1 =2·1+0·(−1)=2. (either no or infinite number). Is A invertible in this case?
(c) Then for what value of k, the system has infinite number of
The second column of C is −2 . solution. Express the resulting solution.
2
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Problems and Solutions Problems and Solutions
Solution Solution
1 −1 2 1 (c) To have infinite number of solution, k = 6, then the
(a) The augmented matrix is 1 1 4 5 1 0 3 3
2 0 h k augmented matrix is 0 1 1 2 .
(b) To find the reduced form 0 0 0 0
So the system of linear equations is transformed to
1 −1 2 1 −R1+R2→R2 1 −1 2 1 R /2→R (
−2R +R →R 2 2
1 1 4 5 1 3 3 0 2 2 4
−→ −→ x +3x =3
1 3
2 0 h k 0 2 h−4 k−2 x +x =2
2 3
1 −1 2 1 R2+R1→R1 1 0 3 3 If we set x = t as the free variable, the solution becomes
0 1 1 2 −2R2+R3→R3 0 1 1 2 3
0 2 h−4 k−2 −→ 0 0 h−6 k−6
x 3 −3
1
x = 2 +t −1 .
if h = 6, the last equation is 0 = k − 6, where there is either 2
x 0 1
infinite number of or no solutions. A is NOT invertible in this 3
case.
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Problems and Solutions Problems and Solutions
Solution - Part 1
Example
−3R +R →R
Find the solution x , x x and x for the following system: 1 0 −1 2 0 1 2 2 1 0 −1 2 0
1 2 3 4 −R +R →R
3 −4 2 0 13 1 3 3 0 −4 5 −6 13
−2R +R →R
1 4 4
x −x +2x =0 −→
1 3 4 1 1 0 −3 −1 0 1 1 −5 −1
3x −4x +2x =13 2 1 −1 −5 −1 0 1 1 −9 −1
1 2 3
x +x −3x =−1
1 2 4
2x +x −x −5x =−1 1 0 −1 2 0 −R2+R3→R3 1 0 −1 2 0
1 2 3 4 R ↔R 0 1 1 −9 −1 4R +R →R 0 1 1 −9 −1
2 4 2 4 4
−→ 0 1 1 −5 −1 −→ 0 0 0 4 0
0 −4 5 −6 13 0 0 9 −42 9
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Problems and Solutions Problems and Solutions
Solution - Part 2
R Solution - Part 2
3 →R3 1 0 −1 2 0 1 0 −1 2 0
4
R
4 →R4 0 1 1 −9 −1 R↔R 0 1 1 −9 −1
9 3 4 1 0 0 0 1
−→ −→ 14 −R +R →R
0 0 0 1 0 0 0 1 − 1 3 2 2
3 R +R →R 0 1 0 0 −2
14 3 1 1
0 0 1 −3 1 0 0 0 1 0 −→ 0 0 1 0 1
14 0 0 0 1 0
R +R →R
3 4 3 3 1 0 −1 0 0
9R +R →R
4 2 2 Therefore, x = 1, x = −2, x = 1, x = 0.
−2R +R →R 0 1 1 0 −1 1 2 3 4
4 1 1
−→ 0 0 1 0 1
0 0 0 1 0
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
Problems and Solutions Problems and Solutions
Example Example
Which of the following matrices is in reduced form? An economy is based on two sectors, energy (E) and water (W).
To produced one dollar’s worth of E requires 0.4 dollar’s worth of E
1 2 0 0 1 and 0.2 dollar’s worth of W, and to produce one dollar’s worth of
1 2 0 0 1
0 0 1 0 0 Wrequires 0.1 dollaar’s worth of E and 0.3 dollar’s worth of W.
A= 0 1 0 0 1 , B= ,
0 0 0 1 1 0 0 0 1 0 (a) Find the technology matrix M for the economy.
0 0 0 0 0 (b) Find the total output for each sector that is needed to satisfy
1 0 0 a final demand of $40 billion for energy and $30 billion for
C =0 1 0. water.
0 0 2
Answers may be found somewhere in the neighbouring slides Answers to the question in the previous slide: B.
Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices Maosheng Xiong Department of Mathematics, HKUST MATH1003 Review: Part 2. Matrices
no reviews yet
Please Login to review.
