# Linear Algebra. Are my methods and solutions correct?

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?

ehild
Homework Helper
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?

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

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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?

ehild
Homework Helper
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?

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

$$\begin{pmatrix} 3 & -1 & 3 \\ 6 & 0 & λ \\ 2 & 1 & 7 \end{pmatrix}$$

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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ehild
Homework Helper
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