Find a number t such that (3, 1, 4), (2, -3, 5), (5, 9, y) is not linearly independent in R^3

To find a value of \( t \) such that the vectors \( (3, 1, 4) \), \( (2, -3, 5) \), and \( (5, 9, t) \) are **not linearly independent** in \( \mathbb{R}^3 \), we need to determine when these vectors become linearly dependent.

### Step-by-step solution:

1. **Linear dependence condition**:
   Vectors are linearly dependent if there exist constants \( c_1, c_2, c_3 \), not all zero, such that:
   \[
   c_1 \cdot (3, 1, 4) + c_2 \cdot (2, -3, 5) + c_3 \cdot (5, 9, t) = (0, 0, 0)
   \]
   In matrix form, this becomes:
   \[
   \begin{pmatrix}
   3 & 2 & 5 \\
   1 & -3 & 9 \\
   4 & 5 & t
   \end{pmatrix}
   \]
   We are looking for values of \( t \) such that the determinant of this matrix is zero, because the determinant of the coefficient matrix being zero implies linear dependence.

2. **Determinant calculation**:
   The determinant of the matrix is given by:
   \[
   \det \begin{pmatrix}
   3 & 2 & 5 \\
   1 & -3 & 9 \\
   4 & 5 & t
   \end{pmatrix}
   = 3 \cdot \det \begin{pmatrix} -3 & 9 \\ 5 & t \end{pmatrix}
   - 2 \cdot \det \begin{pmatrix} 1 & 9 \\ 4 & t \end{pmatrix}
   + 5 \cdot \det \begin{pmatrix} 1 & -3 \\ 4 & 5 \end{pmatrix}
   \]

   Let's compute these 2x2 determinants step by step.

   - \( \det \begin{pmatrix} -3 & 9 \\ 5 & t \end{pmatrix} = (-3)(t) - (9)(5) = -3t - 45 \)
   - \( \det \begin{pmatrix} 1 & 9 \\ 4 & t \end{pmatrix} = (1)(t) - (9)(4) = t - 36 \)
   - \( \det \begin{pmatrix} 1 & -3 \\ 4 & 5 \end{pmatrix} = (1)(5) - (-3)(4) = 5 + 12 = 17 \)

3. **Substitute the results into the determinant expression**:
   \[
   \det = 3 \cdot (-3t - 45) - 2 \cdot (t - 36) + 5 \cdot 17
   \]
   Simplifying each term:
   \[
   = 3(-3t - 45) = -9t - 135
   \]
   \[
   = -2(t - 36) = -2t + 72
   \]
   \[
   = 5 \cdot 17 = 85
   \]
   Now combine all the terms:
   \[
   \det = (-9t - 135) + (-2t + 72) + 85
   \]
   Simplify further:
   \[
   \det = -9t - 135 - 2t + 72 + 85 = -11t + 22
   \]

4. **Set the determinant equal to zero**:
   For the vectors to be linearly dependent, the determinant must be zero:
   \[
   -11t + 22 = 0
   \]
   Solving for \( t \):
   \[
   t = \frac{22}{11} = 2
   \]

### Conclusion:
The vectors \( (3, 1, 4) \), \( (2, -3, 5) \), and \( (5, 9, t) \) are **linearly dependent** when \( t = 2 \).

Would you like any more details on this, or do you have further questions?

---

Here are 5 follow-up questions:
1. What is the geometrical interpretation of linear dependence in \( \mathbb{R}^3 \)?
2. How can we determine if more than three vectors are linearly dependent in \( \mathbb{R}^3 \)?
3. What happens to the determinant when the vectors are linearly independent?
4. How do you interpret the determinant of a 2x2 matrix geometrically?
5. What is the relationship between the rank of a matrix and the linear dependence of its rows or columns?

**Tip**: In \( \mathbb{R}^3 \), if three vectors are linearly dependent, they all lie in the same plane.

Find a number t such that (3, 1, 4), (2, -3, 5), (5, 9, y) is not linearly independent in R^3.

Learn how to find a value of t such that the vectors (3, 1, 4), (2, -3, 5), and (5, 9, t) in R^3 become linearly dependent. This problem involves calculating the determinant of a 3x3 matrix and applying the linear dependence theorem. Suitable for undergraduate students studying linear algebra.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation