Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 1 / 8C / a / Solution

Adding the three equations for $s,t$ gives the necessary solvability condition $a+b+c=3$. Conversely, subtraction of the first two and use of the third forceIf $a+b+c=3$, these values also satisfy the first two equations: for example,The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:

For the three-variable system the matrix and data vector arePerform Gaussian elimination on the augmented matrix. The operations $R_2\leftarrow 5R_2-2R_1$ and $R_3\leftarrow 5R_3+R_1$ giveEach operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow 7R_3+R_2$ givesBack substitution now yieldsand thenAll three pivots are nonzero. Thus the solution exists uniquely for every $s,t$, the matrix rank is $3$, and the kernel of a linear map represented by $A$ is $\{0\}$. The general rank-nullity theorem for a matrix with $m$ columns ishere $m=3$.

Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly whenWhen this holds, their unique values are $s=\frac{3}{2}(x-y)$ and $t=\frac{1}{2}(1+x+y-8z)$. The allowable triples form an affine plane, not all of ${\mathbb{R}}^3$.