Testing Vectors for Linear Independence in ℝ³
One of the central questions in linear algebra is whether a given set of vectors is linearly independent. If they are, no vector in the set can be written as a combination of the others — each one contributes a genuinely new direction. If they are linearly dependent, at least one is redundant.
Let's work through a concrete example with three vectors in ℝ³ and see how the determinant gives us a clean answer.
The Problem
Determine whether the vectors v₁ = (1, −2, 3), v₂ = (5, 6, −1), and v₃ = (3, 2, 1) are linearly independent or linearly dependent in ℝ³.
Setting Up the Linear Combination
We look for scalars x, y, z — not all zero — such that the vectors sum to the zero vector. If only the trivial solution x = y = z = 0 exists, the vectors are independent.
x(1, −2, 3) + y(5, 6, −1) + z(3, 2, 1) = (0, 0, 0)
The System of Equations
Equating each component gives us three linear equations:
−2x + 6y + 2z = 0
3x − y + z = 0
Matrix Form
We write this as the homogeneous system Ax = 0, where the columns of A are our three vectors:
−2 6 2
3 −1 1
y
z
0
0
A homogeneous system always has the trivial solution. The key question is whether non-trivial solutions exist — and that hinges on the determinant of A.
Computing the Determinant
Expanding along the first row using cofactor expansion:
= 1·(6 + 2) − 5·(−2 − 6) + 3·(2 − 18)
= 8 + 40 − 48
= 0
Interpretation
A determinant of zero means the matrix A is singular — it has no inverse, and the system Ax = 0 has infinitely many solutions beyond just the trivial one. In geometric terms, the three vectors are coplanar: they all lie within the same plane through the origin and cannot span all of ℝ³.
Since |A| = 0, a non-trivial solution exists for the system.
∴ The vectors v₁, v₂, and v₃ are linearly dependent.
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