Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 1 / 6B (Vectors and Matrices)

For any linear system $Ax=c$, a solution exists exactly when $c$ belongs to the image of a linear map represented by $A$. If $x_0$ is one solution, all others are exactlysince $A(x-x_0)=0$, and conversely adding a kernel vector preserves the equation. Thus there are no solutions if $c\notin \mathrm{im}A$, one if $c\in \mathrm{im}A$ and $\ker A=\{0\}$, and infinitely many if $c\in \mathrm{im}A$ and the kernel is nontrivial: every real multiple of a nonzero kernel vector gives a distinct solution. Any two solutions differ by a kernel vector. Equivalently, consistency means the augmented matrix and $A$ have equal rank of a matrix; a consistent square linear system is unique exactly when its determinant is nonzero.

For the given matrix, set $d=b-a$ and $s=2a+b$. Perform the column operation $C_1\leftarrow C_1+C_2+C_3$, then the row operations $R_2\leftarrow R_2-R_1$ and $R_3\leftarrow R_3-R_1$. Factoring the first column givesThese operations are determinant-preserving, and the displayed polynomial identity remains valid even when a factored expression vanishes.

To find all kernels and images of a linear map, putIf $J$ is the all-ones matrix, then $A=aJ+dP$. The direct sum ${\mathbb{R}}^3=\mathbb{R}u\oplus H$ is invariant under $A$: on $\mathbb{R}u$ it acts by multiplication by $s$, and on $H$ by $dP$. Since $P^3=I$, its restriction to $H$ is invertible.

For $d\ne 0$ and $s\ne 0$, the kernel is $\{0\}$ and the image of a linear map is all of ${\mathbb{R}}^3$. Decompose the right-hand side asApplying the inverse on each invariant summand gives the unique solutionFor $d\ne 0$ and $s=0$, the kernel is $\mathbb{R}u$ and the image of a linear map is $H$. A solution exists exactly when $c+2=0$. For $c=-2$, all solutions areFor $d=0$ and $a\ne 0$, so $a=b$, the kernel is $H$ and the image of a linear map is $\mathbb{R}u$. A solution exists exactly when $c=1$. All such solutions areFinally, for $a=b=0$, the kernel is ${\mathbb{R}}^3$ and the image of a linear map is $\{0\}$. The right-hand side has first entry one, so there is no solution for any $c$. These four cases exhaust all parameter values.