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Consider the following set of four vectors {a1,. ,as} and the vector b. a = |2 az = 0 and b = 1. Solve the vector equation r¡aj + xza2 + 13â3 + L4ās b. Assume that the a, vector is removed from the set. 2. Explain whether the vector equation r,a¡ + 1,az + 1zāz = b can also be solved. PROBLEM 1. Solution The first problem is solved by row reduction of the augmented matrix 1 -1 3 1| 4 [ A|b] =| 2 1 0 -4 10 1 -1 10 1 0 1 -2 -2|-3 0 0 0 0 -1 The system is consistent and contains two free variables, x3 and 14. The general solution is written as I3 It can be seen from the row reduced matrix above that a4 = -a1 – 2az i.e. a4 is linearly dependent on aj and a2. Therefore, span{aį,…., a}=span{a,…..., as} and the vector equation can still be solved with a, removed. In fact, span{a,…..., a}=span{a,, a2} and az can also be removed.

Consider the following set of four vectors {a1,. ,as} and the vector b. a = |2 az = 0 and b = 1. Solve the vector equation r¡aj + xza2 + 13â3 + L4ās b. Assume that the a, vector is removed from the set. 2. Explain whether the vector equation r,a¡ + 1,az + 1zāz = b can also be solved. PROBLEM 1. Solution The first problem is solved by row reduction of the augmented matrix 1 -1 3 1| 4 [ A|b] =| 2 1 0 -4 10 1 -1 10 1 0 1 -2 -2|-3 0 0 0 0 -1 The system is consistent and contains two free variables, x3 and 14. The general solution is written as I3 It can be seen from the row reduced matrix above that a4 = -a1 – 2az i.e. a4 is linearly dependent on aj and a2. Therefore, span{aį,…., a}=span{a,…..., as} and the vector equation can still be solved with a, removed. In fact, span{a,…..., a}=span{a,, a2} and az can also be removed.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Hello bartleby

i marked the question i dont understand. the solution beneth is my teachers answer. however i dont understand whats really meant. i hope you could explain it more simple.

PROBLEM 1. Consider the following set of four vectors {a1,. a4} and the vector b. ..... [3 a = 2 , a2 and b = dz = a = 1. Solve the vector equation xịaj + x2a2 + x3a3 + x4ąs = b. Assume that the a, vector is removed from the set. 2. Explain whether the vector equation ,a + X2a, + ¤zâz b can also be solved. PROBLEM 1. Solution The first problem is solved by row reduction of the augmented matrix 1 -1 3 0 -4 1 0 1 -1 10 1 0 1 -2 -2|-3 0 0 0 1 4 -1 1 [ A|b] = | 2 1 The system is consistent and contains two free variables, rz and a4. The general solution is written as 2 + I3 + x4 I3 1 It can be seen from the row reduced matrix above that a4 = dependent on a1 and a2. Therefore, span{a1,..., a}=span{a1,..., a3} and the vector equation can still be solved with a, removed. In fact, span{a,..., aj}=span{a1, a2} and az can also be removed. 2az i.e. a4 is linearly
Transcribed Image Text:PROBLEM 1. Consider the following set of four vectors {a1,. a4} and the vector b. ..... [3 a = 2 , a2 and b = dz = a = 1. Solve the vector equation xịaj + x2a2 + x3a3 + x4ąs = b. Assume that the a, vector is removed from the set. 2. Explain whether the vector equation ,a + X2a, + ¤zâz b can also be solved. PROBLEM 1. Solution The first problem is solved by row reduction of the augmented matrix 1 -1 3 0 -4 1 0 1 -1 10 1 0 1 -2 -2|-3 0 0 0 1 4 -1 1 [ A|b] = | 2 1 The system is consistent and contains two free variables, rz and a4. The general solution is written as 2 + I3 + x4 I3 1 It can be seen from the row reduced matrix above that a4 = dependent on a1 and a2. Therefore, span{a1,..., a}=span{a1,..., a3} and the vector equation can still be solved with a, removed. In fact, span{a,..., aj}=span{a1, a2} and az can also be removed. 2az i.e. a4 is linearly
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