Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 1 / 2B (Algebra and Geometry)

The coefficient matrix has determinantA square linear system has a unique solution for every right-hand side precisely when its coefficient matrix is an invertible matrix. Here a unique solution exists exactly when $p\ne 1$, for every real $t$.

For an explicit Gaussian elimination, subtract twice the second equation from the third, then use the second equation, obtaining $x=2z+2t-t^2$ and $y=(t^2-t)/2-3z$. The first equation determinesand the preceding formulas determine $x,y$. When $p=1$ this last coefficient vanishes; the next parts determine compatibility rather than dividing by zero.

For $p=1$, the third coefficient row equals three times the second minus twice the first. The first two rows are independent, so the coefficient matrix rank is two. The same dependence must hold for the right-hand side, givingThus infinitely many solutions occur exactly for $p=1$ and $t\in \{1,2\}$. The null space has dimension one by the rank-nullity theorem, and an explicit parameterization, with $s\in \mathbb{R}$, isSubstitution verifies all three equations in each family.

When $p=1$ but $t^2\ne 3t-2$, 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 $p\ne 1$ part (i) already supplies a unique solution. Consequently there is no solution exactly when $p=1$ and $t\notin \{1,2\}$.