Math Problem Statement

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

Solution

To find a value of tt such that the vectors (3,1,4)(3, 1, 4), (2,3,5)(2, -3, 5), and (5,9,t)(5, 9, t) are not linearly independent in R3\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 c1,c2,c3c_1, c_2, c_3, not all zero, such that: c1(3,1,4)+c2(2,3,5)+c3(5,9,t)=(0,0,0)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:

    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:

    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(3t45)2(t36)+517\det = 3 \cdot (-3t - 45) - 2 \cdot (t - 36) + 5 \cdot 17 Simplifying each term: =3(3t45)=9t135= 3(-3t - 45) = -9t - 135 =2(t36)=2t+72= -2(t - 36) = -2t + 72 =517=85= 5 \cdot 17 = 85 Now combine all the terms: det=(9t135)+(2t+72)+85\det = (-9t - 135) + (-2t + 72) + 85 Simplify further: det=9t1352t+72+85=11t+22\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=0-11t + 22 = 0 Solving for tt: t=2211=2t = \frac{22}{11} = 2

Conclusion:

The vectors (3,1,4)(3, 1, 4), (2,3,5)(2, -3, 5), and (5,9,t)(5, 9, t) are linearly dependent when t=2t = 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 R3\mathbb{R}^3?
  2. How can we determine if more than three vectors are linearly dependent in R3\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 R3\mathbb{R}^3, if three vectors are linearly dependent, they all lie in the same plane.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Linear Dependence
Determinants

Formulas

Determinant of a 3x3 matrix
Linear dependence condition: c1*v1 + c2*v2 + c3*v3 = 0

Theorems

Linear Dependence Theorem
Matrix Determinant Theorem

Suitable Grade Level

Undergraduate Level (Linear Algebra)