R is the reduced row echelon form of a real 4×4 A, and its third column is (0,1,0,0)ᵀ. P:if (α,β,γ,0)ᵀ solves Ax=0, then γ=0. Q:rank[A|b]=rank[R|b] for every b∈ℝ⁴. Choose truth values.
P fails for $A=R=\begin{pmatrix}1&0&0&0\\0&1&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$: (0,−1,1,0) solves Ax=0 with γ=1. Q fails because b must undergo the same row operations. For example take A with first row (1,0,0,0), third row (0,1,1,0), and other rows zero. Its R is the displayed matrix. For b=e₃, rank[A|b]=2 but rank[R|b]=3. **D**.
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