Question:

<p><strong>Step 1:</strong> Write the augmented matrix and find when rank(A) ≠ rank(A|b).</p><p>The coefficient matrix is:<br/>$$\begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix}$$</p><p><strong>Step 2:</strong> Expand the determinant:<br/>$$R_2 - R_1: (0, 2-a, 1)$$<br/>$$R_3 - R_1: (0, 5-a, 2)$$<br/>$$\det(A) = 1[(2-a)·2 - (5-a)·1] = 1[4-2a-5+a] = -1-a$$</p><p>For no solution, we need det(A) = 0, so: $a = -1$</p><p><strong>Step 3:</strong> When $a = -1$, the first two equations become dependent. Row reduce with $a = -1$:<br/>Row 2: $x - y + 2z = 6$ becomes dependent on Row 1: $x - y + z = 3$<br/>This gives $z = 3$</p><p><strong>Step 4:</strong> Substitute $a = -1, z = 3$ into Row 3:<br/>$$x + 5y + 3(3) = b$$<br/>$$x + 5y + 9 = b$$<br/>But from Row 1: $x - y + 3 = 3$, so $x = y$<br/>Thus: $y + 5y + 9 = b$ → $6y + 9 = b$</p><p><strong>Step 5:</strong> For no solution, Row 3 must contradict the others. This happens when $b ≠ 9$ (when $y$ cannot be determined consistently).<br/>More directly: rank(A) = 2 when $a = -1$, but rank(A|b) = 3 when $b ≠ 9$.</p><p><strong>Correct condition: </strong>$a = -1$ and $b ≠ 9$</p><p>∴ Answer: B</p>