Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 2 / c

A system of linear equations $T_{\alpha }x=b$ is solvable exactly when $b\in \mathrm{im}{T}_{\alpha }$. If $x_0$ is one solution, thenso its solution set is $x_0+\ker T_{\alpha }$, a translate of the kernel of a linear map.

For $\alpha \ne 2$, the matrix is an invertible matrix, andEvery $b$ is in the image of a linear map and has exactly one preimage. For $\alpha =2$, its first two columns span the image of a linear map, while the corresponding homogeneous linear system gives the kernel of a linear map:The prescribed $b=(-1,\beta ,1{)}^T$ satisfies the displayed condition precisely when $\beta =0$. If $\beta \ne 0$, it lies outside the image of a linear map and there is no solution. If $\beta =0$, it lies in the image of a linear map, and the nonzero kernel of a linear map supplies infinitely many solutions.

For $\alpha \ne 2$, $A^{\prime }=I_3$, whose row space is ${\mathbb{R}}^3=\mathrm{im}{T}_{\alpha }$. For $\alpha =2$, elementary row operations give the reduced row echelon formIts row space is $\{(u,v,w):u+v+w=0\}$, whereas the image of a linear map is $\{(u,v,w):u-v+w=0\}$. These differ: $(0,1,-1)$ belongs to the former but not the latter. Thus (A) is true exactly when $\alpha \ne 2$. Rows are being identified with vectors in ${\mathbb{R}}^3$ in this comparison.

Every elementary column operation preserves the linear span of the columns: each new column is a linear combination of the old columns, and the inverse operation supplies the reverse inclusion. Thus the column space of $A^{\prime \prime }$ equals that of $A$, which is precisely the image of a linear map $T_{\alpha }$. Consequently (B) is true for every $\alpha \in \mathbb{R}$.

For $\alpha \ne 2$, both reduced forms are $I_3$, so $A^{\prime \prime }=(A^{\prime }{)}^T$. At $\alpha =2$, the reduced column echelon form is the transpose of the reduced row echelon form of $A^T$, givingFor example, $C_2\mapsto -C_2$ followed by $C_3\mapsto C_3+C_1-C_2$, using the updated second column, produces the displayed $A^{\prime \prime }$ directly. Its third column is zero. The two matrices differ, so (C) is true exactly when $\alpha \ne 2$. Taking a transpose exchanges the kind of reduction, but also replaces the original $A$ by $A^T$; it does not generally identify the two reduced forms of $A$ itself.