Discussion
Let \(A\) be a \(5 \times 3\) real matrix of rank 2 and \(b\) be a non-zero \(5 \times 1\) real column vector. Suppose \(\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\) and \(\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}\) are two solutions of the system of linear equations \(Ax = b\).
Which ONE of the following is also a solution of \(Ax = b\)?
Answer: D
The other solution is \(\begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix}\).
\(A\) has 3 columns and rank 2, so its null space has dimension \(3 - 2 = 1\). The difference of the two given solutions lies in it:
\[ \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix} \]
so the null space is spanned by \((1,1,1)^{T}\) and every solution has the form \((1,2,3)^{T} + s(1,1,1)^{T}\). Taking \(s = 6\) gives \((7,8,9)^{T}\).
\((3,3,3)^{T}\) solves \(Ax = 0\), not \(Ax = b\), since \(b \neq 0\). \((5,7,9)^{T}\) is the sum of the two solutions, so \(A\) sends it to \(2b \neq b\). \((5,6,9)^{T}\) differs from \((1,2,3)^{T}\) by \((4,4,6)^{T}\), which is not a multiple of \((1,1,1)^{T}\).
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