Question:

<p><strong>Step 1:</strong> Write the augmented matrix:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 2 & 3 & 2 & | & 5 \\ 2 & 3 & a^2-1 & | & a+1 \end{bmatrix}$$</p><p><strong>Step 2:</strong> Apply row operations: $R_2 - 2R_1$ and $R_3 - 2R_1$:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & 0 & | & 1 \\ 0 & 1 & a^2-3 & | & a-3 \end{bmatrix}$$</p><p><strong>Step 3:</strong> Apply $R_3 - R_2$:</p><p>$$\begin{bmatrix} 1 & 1 & 1 & | & 2 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & a^2-3 & | & a-3 \end{bmatrix}$$</p><p><strong>Step 4:</strong> For consistency, if $a^2 - 3 = 0$, then $a - 3$ must also equal 0. But $a^2 = 3$ gives $a = ±\sqrt{3}$, while $a = 3$ is separate. If $a^2 - 3 \neq 0$, the system is always consistent. For the system to have a solution when $a^2 - 3 = 0$, we need $a - 3 = 0$, which is impossible since $a = ±\sqrt{3} \neq 3$.</p><p><strong>Step 5:</strong> Therefore, the system has a solution when $a^2 - 3 \neq 0$, i.e., $a \neq ±\sqrt{3}$, OR when both coefficients are zero (impossible here). The condition is: <strong>$a \neq ±\sqrt{3}$ or equivalently $a^2 \neq 3$</strong></p><p>∴ Answer: C</p>