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

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.

Writing the unknown vector as $(x,y,z{)}^T$, the first two linear equations give $x=z-1$ and $y=z-\beta$. Substitution into the third givesThus, when $\alpha \ne 2$, put $t=-\beta /(\alpha -2)$. There is exactly one solution:When $\alpha =2$, the remaining equation is $0=-\beta$. HenceThe second case is an affine line in a vector space, so it contains infinitely many solutions. These cases exhaust all real numbers $\alpha ,\beta$.

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.