Solution
= Solution
When $p=1$ but $t^2\ne3t-2$, the dependent coefficient rows would require an impossible equality between their right-hand sides. Thus the <augmented matrix> has larger <matrix rank> than the coefficient <matrix>, and the <linear system> is inconsistent. For $p\ne1$ part (i) already supplies a unique solution. Consequently \b[there is no solution exactly when $p=1$ and $t\notin\{1,2\}$].