Science:Math Exam Resources/Courses/MATH221/December 2008/Question 01 (b)
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) • Q8 (a) • Q8 (b) • Q9 (a) • Q9 (b) • Q10 (a) • Q10 (b) •
Question 01 (b) 

Let denote the matrix Determine whether the vector is in the column space of the matrix . 
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? 
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

For to be in the column space of , it must be a linear combination of the basis vectors. From part (a) we know that a basis for the column space of consists of the first two column vectors of . Hence, check if there are constants , such that 
Hint 2 

Rewriting the equation in the hint above: If was in the column space of , how many solutions are there to the equation If we know from part (a) that there is only the trivial solution. Otherwise, divide this equation by and set , to get back to the equation on the first hint. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

Solution 1 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. To determine if b is in the column space of A, we have to make sure that b is a linear combination of the column space of A If there is a solution for , then we can write b as a linear combination of the columns of A We can do this by row reduce the matrix [Ab] The last row of the reduced row form shows that there is no solution for this augmented matrix. Therefore, vector b = is not in the column space of matrix A = 
Solution 2 

Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. The fastest way is to check if the matrix, whose first two columns are the basis of the column space of and the third column is the vector , has determinante . Thereby we check if these three vectors are linear dependent. If yes, then lies in the column space of .
Hence, the vectors are linear independent and does not lies in the column space of . 