Let's go through the questions visible from the image and provide solutions:

Question 5:

Determine whether the vectors
$\mathbf{v}_1 = (1, -2, 3), \ \mathbf{v}_2 = (5, 6, -1), \ \mathbf{v}_3 = (3, 2, 1)$ form a linearly dependent or linearly independent set in $\mathbb{R}^3$.

Solution:

To check if the set of vectors $\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is linearly dependent or independent, we need to see if the only solution to the equation $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + c_3 \mathbf{v}_3 = \mathbf{0}$ is $c_1 = c_2 = c_3 = 0$. This boils down to checking if the determinant of the matrix formed by these vectors as columns is zero.

The matrix of the vectors is: $A = \begin{pmatrix} 1 & 5 & 3 \\ -2 & 6 & 2 \\ 3 & -1 & 1 \end{pmatrix}$ Now, calculate the determinant of $A$: $\det(A) = 1 \cdot \begin{vmatrix} 6 & 2 \\ -1 & 1 \end{vmatrix} - 5 \cdot \begin{vmatrix} -2 & 2 \\ 3 & 1 \end{vmatrix} + 3 \cdot \begin{vmatrix} -2 & 6 \\ 3 & -1 \end{vmatrix}$ This evaluates to: $\det(A) = 1 \cdot (6 - (-2)) - 5 \cdot (-2 - 6) + 3 \cdot (-2 \cdot -1 - 3 \cdot 6)$ $\det(A) = 1 \cdot 8 - 5 \cdot (-8) + 3 \cdot (2 - 18)$ $\det(A) = 8 + 40 + 3 \cdot (-16) = 48 - 48 = 0$

Since the determinant is zero, the vectors are linearly dependent.

Question 6:

If the vectors
$(0, 1, a), \ (1, a, 1), \ (a, 1, 0)$ are linearly dependent, find the value of $a$.

Solution:

For the vectors to be linearly dependent, the determinant of the matrix formed by these vectors as columns must be zero.

The matrix is: $A = \begin{pmatrix} 0 & 1 & a \\ 1 & a & 1 \\ a & 1 & 0 \end{pmatrix}$ Now, calculate the determinant: $\det(A) = 0 \cdot \begin{vmatrix} a & 1 \\ 1 & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ a & 0 \end{vmatrix} + a \cdot \begin{vmatrix} 1 & a \\ a & 1 \end{vmatrix}$ This simplifies to: $\det(A) = 0 - 1 \cdot (1 \cdot 0 - a \cdot 1) + a \cdot (1 \cdot 1 - a \cdot a)$ $\det(A) = a + a(1 - a^2) = a + a - a^3 = 2a - a^3$ For linear dependence, set the determinant to zero: $2a - a^3 = 0$ Factor this equation: $a(2 - a^2) = 0$ Thus, $a = 0$ or $a^2 = 2 \Rightarrow a = \pm \sqrt{2}$.

So, the values of $a$ that make the vectors linearly dependent are: $a = 0, \ a = \sqrt{2}, \ a = -\sqrt{2}.$

Question 7:

If $W_1, W_2$ are two subspaces of a finite-dimensional vector space $V$ with: $\dim(W_1) = 2, \ \dim(W_2) = 3, \ \dim(W_1 + W_2) = 4$ find $\dim(W_1 \cap W_2)$.

Solution:

We can use the following formula that relates the dimensions of the sum and intersection of two subspaces: $\dim(W_1 + W_2) = \dim(W_1) + \dim(W_2) - \dim(W_1 \cap W_2)$ Substitute the known values: $4 = 2 + 3 - \dim(W_1 \cap W_2)$ This simplifies to: $4 = 5 - \dim(W_1 \cap W_2)$ So, $\dim(W_1 \cap W_2) = 1.$

Would you like further details or have any questions on these problems?

Here are some related questions:

1. How can you generalize the concept of linear independence to an arbitrary number of vectors in $\mathbb{R}^n$?
2. What role do determinants play in solving systems of linear equations?
3. What is the significance of the intersection of subspaces in vector space theory?
4. How do the properties of subspaces affect the solution of a system of linear equations?
5. Can two non-zero vectors in $\mathbb{R}^3$ ever be linearly independent? Why or why not?

Tip: Always check if a set of vectors is linearly dependent by calculating the determinant of the matrix they form. If the determinant is zero, the vectors are dependent.