Exercise 3: Confirming a Basis for the 2D Plane — Possible Solution
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GIVEN
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m = [2, 5]
n = [-1, 3]

STEP 1: THE DETERMINANT TEST
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det([m, n]) = (2)(3) - (5)(-1) = 6 + 5 = 11

Since 11 is nonzero, m and n are linearly independent, per this
chapter's own Chapter 7 determinant shortcut.

STEP 2: CONFIRMING {m, n} IS A BASIS
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Per this chapter's own definition, a basis must be BOTH linearly
independent AND span the whole space. {m, n} is already confirmed
independent (Step 1). Per this chapter's own worked example, any two
non-parallel vectors in 2D automatically span the entire plane -
since m and n are independent (hence non-parallel), their span is all
of R^2. Both conditions are satisfied, so {m, n} is a valid basis for
the 2D plane.

WHY EXACTLY TWO INDEPENDENT VECTORS ARE NECESSARY AND SUFFICIENT
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The 2D plane has dimension exactly 2 (per this chapter's own
dimension fact, every basis for a given space has the same vector
count) - one independent vector can only ever span a single line, too
few directions to reach every point in the plane, while any third
vector added to two already-independent 2D vectors is guaranteed
redundant (its own span-collapsing row-reduction argument, applied
generally), so exactly two is the minimum needed to reach everywhere
and the maximum that can stay non-redundant at the same time.

WHY THIS WORKS AS AN ANSWER
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The determinant test confirms independence using this chapter's own
formula, the basis conclusion combines that result with this
chapter's own span table finding for non-parallel 2D vectors, and the
"why exactly two" explanation is grounded directly in this chapter's
own dimension fact rather than treated as a separate, unrelated
observation.