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Can Three Matrices in M₂ₓ₃(ℝ) Be Linearly Dependent? A Complete Row-Reduction Proof

Linear independence isn't just for column vectors — it applies to any vector space, including spaces of matrices. In this post, we take three 2×3 matrices, stack their entries into a 6×3 coefficient system, and use row reduction to determine their rank. If rank < 3, the set is dependent. Let's find out.  

In the vector space M₂ₓ₃(ℝ) — the set of all real 2×3 matrices — linear independence means exactly what it does for ordinary vectors: no matrix in the set can be expressed as a scalar combination of the others. The test is identical too: set up a homogeneous linear combination equal to the zero matrix, extract the system of equations, and analyse it.

The Problem3

Example —3

Verify whether the set S = {A₁, A₂, A₃} in M₂ₓ₃(ℝ) is linearly dependent or not, where:

Setting Up the Linear Combination

To test linear dependence, we assume that for some scalars α, β, γ ∈ ℝ the following holds:

αA₁ + βA₂ + γA₃ = O

Expanding this explicitly:

α·( 1 -3 2 / -4 0 5) + β·(-3 7 4 / 6 -2 -7) + γ·(-2 3 11 / -1 -3 2) = (0 0 0 / 0 0 0)

Extracting the System of Equations

Equating the six corresponding entries (two rows × three columns) of both sides gives a homogeneous system of 6 equations in 3 unknowns:

α − 3β − 2γ = 0      −3α + 7β + 3γ = 0      2α + 4β + 11γ = 0      −4α + 6β − γ = 0      0α−2β − 3γ = 0      5α − 7β + 2γ = 0     

We write the coefficient matrix A (6×3) and row-reduce it. If rank(A) < 3, a non-trivial solution exists and the set is linearly dependent.

Row Reduction — Step by Step

Initial Matrix A

1 -3 -2 -3 7 3 2 4 11 -4 6 -1 0 -2 -3 5 -7 2

R₂ ← R₂ + 3R₁ R₃ ← R₃ − 2R₁ R₄ ← R₄ + 4R₁ (R₅ unchanged) R₆ ← R₆ − 5R₁

↓ After Row Operations

After eliminating column 1

1 -3 -2 0 -2 -3 0 10 15 0 -6 -9 0 -2 -3 0 8 12

R₂ ← −½ · R₂

↓ Scale pivot row

After scaling R₂

1 -3 -2 0 1 3/2 0 10 15 0 -6 -9 0 -2 -3 0 8 12

R₃ ← R₃ − 10R₂ R₄ ← R₄ + 6R₂ R₅ ← R₅ + 2R₂ R₆ ← R₆ − 8R₂

↓ Eliminate column 2

Row Echelon Form (REF)

1 -3 -2 0 1 3/2 0 0 0 0 0 0 0 0 0 0 0 0

← Pivot row 1 ← Pivot row 2 ← zero row ← zero row ← zero row ← zero row

Reading the Result

The row echelon form has exactly 2 non-zero rows, so rank(A) = 2. Since rank(A) = 2 < 3 (the number of unknowns α, β, γ), the system has infinitely many non-trivial solutions. There exist scalars α, β, γ — not all zero — satisfying αA₁ + βA₂ + γA₃ = O.

rank(A) = 2 < 3  →  Non-trivial solution exists

∴ The set S = {A₁, A₂, A₃} is Linearly Dependent in M₂ₓ₃(ℝ)

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