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Homework 1
Math 308
due Oct. 1st, 2008
Ch 1.1, Exercise 1-
Ex 1 linear
Ex 2 non linear (the term x 1 x 2 )
Ex 3 linear (sin^2 x 1 + cos^2 x 1 = 1)
Ex 4 non linear
Ex 5 non linear
Ex 6 linear
Ch 1.1, Exercise 12
x 1
x 2
The graphs of the two equations are two par- allel lines. There is therefore no solution to the system.
Ch 1.1, Exercise 26
The coefficient matrix is
A =
The augmented matrix is
B =
Ch 1.1, Exercise 34
The augmented matrix is
B =
The elementary equation operations E 2 + E 1 and E 3 + 2E 1 transform the system into the following one:
x 1 + x 2 + x 3 − x 4 = 1 2 x 2 = 4 3 x 2 + 3 x 3 − 3 x 4 = 4
The elementary row operations R 2 + R 1 and R 3 + 2R 1 reduces the augmented matrix into the following matrix:
Ch 1.2, Exercise 2, 6
Ex2 a) The matrix is in echelon form.
b) The row operation R 1 − 2 R 2 transforms the matrix to the reduced row echelon form [ 1 0 − 7 0 1 3
]
Ex 6 a) The matrix is not in echelon form
b) The row operation 1 / 2 R 1 transforms the matrix to the echelon form [ 1 0 3 / 2 1 / 2 0 0 1 2
]
Ch 1.2, Exercise 24
The augmented matrix of the system is [ 1 − 1 1 3 2 1 − 4 − 3
]
The row operations R 2 − 2 R 1 and 1 / 3 E 2 transform it in echelon form
[ 1 − 1 1 3 0 1 − 2 − 3
]
then the row operation R 1 + R 2 transforms it to reduced row echelon form
[ 1 0 − 1 0 0 1 − 2 − 3
]
Therefore the system is equivalent to
x 1 − x 3 = 0 x 2 − 2 x 3 = − 3
There is an infinite number of solutions of the form x 1 = x 3 , x 2 = 2x 3 and x 3 arbitrary.
Ch 1.2, Exercise 30
The augmented matrix of the system is
8: A homogeneous system has always at least one solution (the point (x 1 = 0,... , xn = 0). If it has 4 equations and 5 unknowns it necessarily has infinitely many solutions.
12: A homogeneous system of 4 equations in 3 unknowns can either have a unique solution or an infinite number of solutions.
14: A system that has x 1 = − 1 , x 2 = 0,x 3 = 2, and x 4 = − 3 as solution has either a unique solution or infinitely many solutions. Now if it has 3 equations and 4 unknowns, necessarily has infinitely many solutions.
Ch 1.3, Exercise 28
A conic is given by an equation of the type
ax^2 + bxy + cy^2 + dx + ey + f = 0,
where a, b or c is non zero. The parameters a, b, c, d, e, f must satisfy the system
16 a − 4 d + f = 0 4 a + 4 b + 4 c − 2 d − 2 e + f = 0 9 c + 3 e + f = 0 a + b + c + d + e + f = 0 16 a + 4 d + f = 0
The system has infinitely many solutions of the form a = − 161 f ,b = − 14471 f , c = 181 f , d = 0, e = − 21 f , for f arbitrary. Thus an equation for the conic is
9 x^2 + 71xy − 8 y^2 + 72y − 144 = 0.
