Linear Algebra. Are my methods and solutions correct?

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Homework Help Overview

The discussion revolves around three vectors in R3: u1=(3,6,2), u2=(-1,0,1), and u3=(3,λ,7). Participants are exploring conditions under which these vectors span R3 or a plane in R3, as well as the implications of these conditions on the values of λ and k for a related vector v=(8,6,k).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the determinant of the vectors to determine when they span R3 or a plane. There are questions about the correctness of methods used, particularly regarding the implications of the determinant being zero or non-zero for the independence of the vectors.

Discussion Status

Some participants have provided calculations and results for the determinant and values of λ. There is an ongoing examination of the methods used, with requests for verification of the approaches taken. Confusion about the relationship between the determinant values and the spanning conditions has been raised.

Contextual Notes

Participants are grappling with the implications of the determinant being zero versus non-zero, particularly in relation to the independence of the vectors and the conditions for spanning R3 or a plane. There are mentions of specific calculations and results that may need further clarification.

DoderMan
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Consider the following three vectors in R3: u1=(3,6,2) , u2=(-1,0,1) , u3=(3,λ,7)

a) Find all values of λ E R, such that {u1, u2, u3} spans R3, i.e.R3 = span {u1, u2, u3}

b) Find the value of λ E R, such that {u1, u2, u3} spans a plane in R3.

c) Find all values of k E R, such that the vector v=(8,6,k) belongs to the plane spanned by {u1,u2,u3} (for the value of λ which you obtained in part (b))

In part (a) I took the determinante of the vectors {u1,u2,u3} and I got λ=12. Is the procedure and solution correct?

In part (b) I performed Gauss-jorden elimination method on vectors {u1,u2,u3} and I got λ=6. Is the procedure and solution correct?

In part (c) I again performed Gauss-jorden elimination method on vectors {u1,u2,u3,v} and I found that k= (16/3). Is the procedure and solution correct?

I am also confused about part (a) and (b). In part (a) the three vectors span R3 but in part (b) the three vectors span a plane in R3. How is it possible?

Thanks in advance.
 
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DoderMan said:
Consider the following three vectors in R3: u1=(3,6,2) , u2=(-1,0,1) , u3=(3,λ,7)

a) Find all values of λ E R, such that {u1, u2, u3} spans R3, i.e.R3 = span {u1, u2, u3}

b) Find the value of λ E R, such that {u1, u2, u3} spans a plane in R3.

c) Find all values of k E R, such that the vector v=(8,6,k) belongs to the plane spanned by {u1,u2,u3} (for the value of λ which you obtained in part (b))

In part (a) I took the determinante of the vectors {u1,u2,u3} and I got λ=12. Is the procedure and solution correct?

In part (b) I performed Gauss-jorden elimination method on vectors {u1,u2,u3} and I got λ=6. Is the procedure and solution correct?

In part (c) I again performed Gauss-jorden elimination method on vectors {u1,u2,u3,v} and I found that k= (16/3). Is the procedure and solution correct?

I am also confused about part (a) and (b). In part (a) the three vectors span R3 but in part (b) the three vectors span a plane in R3. How is it possible?

Thanks in advance.

What is the determinant when λ=12? The three vectors span the whole 3-dimensional space if they are independent. They span a plane if only two of them are independent.

ehild
 
Last edited:
The determinant is -6 when λ=12. but could you please verify if the methods that i have chosen to solve the problems are correct as well as the answer?

Thanks in advance.
 
DoderMan said:
The determinant is -6 when λ=12. but could you please verify if the methods that i have chosen to solve the problems are correct as well as the answer?

Thanks in advance.
Your "solution" and answers are not correct.
How did you get the values for lambda? What equations did you use? The determinant is an expression including lambda. To get lambda, you need an equation.
The same with the Gauss elimination. You have a matrix. You can transform it into diagonal form. But it is still a matrix. You need an equation to get lambda.
Better to show your derivation in detail.

Anyway, the matrix of the vectors u1,u2,u3

<br /> \begin{pmatrix}<br /> 3 &amp; -1 &amp; 3 \\<br /> 6 &amp; 0 &amp; λ \\<br /> 2 &amp; 1 &amp; 7 <br /> \end{pmatrix}<br />

has determinant D=60-5λ. If λ=12, D=0.


ehild
 
For part (a) I calculated the determinant just like you did. I got D=60-5λ then I put 60-5λ=0 => λ=12.
Then I substituted lambda for 12 and calculated the determinant again and got D= -6

I recalculated part (b) and (c). I took pictures of the calculations and uploaded them here.
In part (b) I got/chose λ=2 and in part (c) I could not find an answer for K.
 

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DoderMan said:
For part (a) I calculated the determinant just like you did. I got D=60-5λ then I put 60-5λ=0 => λ=12.
Then I substituted lambda for 12 and calculated the determinant again and got D= -6

Do you not notice the contradiction? If the determinant is zero when λ=12, how can be the determinant -6 if λ=12?

You made a mistake in the Gauss elimination at the end of the second line, first page.

If the determinant is zero, the equation Ax=0 has non-trivial solutions, that is, linear combination of the column vectors of the matrix can be zero. Are the column vectors independent or dependent then?


ehild
 

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