Math subject GRE Exam 9768 Q.16

Since the vectors other than the one with $m$ are linearly independent, the question is then when these vectors are linearly dependent. This is equivalent to studying the invertibility of the matrix:

$$A(m):= \begin{pmatrix} 1 & 0 & 0 & 1 \\ 2 & 1 & 0 & 1 \\ m & 1 & 0 & 2 \\ 5 & 1 & 1 & 0 \end{pmatrix}$$

It's easy to find the determinant:

$$\det A(m) = -\left| \begin{matrix}1 & 0 & 1 \\ 2 & 1 & 1 \\ m & 1 & 2 \end{matrix} \right| = m-3$$

This shows that a linear combination is possible if and only if $m-3 = 0$ if and only if $m = 3$.

The vectors are $(0,1,1,1)$, $(0,0,0,1)$ and $(1,1,2,0)$. Since your vector has first component $1$, this forces $(1,1,2,0)$ to appear with a coefficient $1$. This reduces your problem to $(1,m-2,5)$ with $(0,0,1)$ and $(1,1,1)$ (forget the first coordinate). The same argument now forces $(1,1,1)$ to appear with coefficient $1$, so $(0,m-3,4)$ is a multiple of $(0,0,1)$, only possible if $m=3$, in which case $(0,0,1)$ appears with coefficient $4$. This gives

$$(1,2,3,5) = (0,1,1,1)+4(0,0,0,1)+(1,1,2,0)$$

The question is the same as asking for what $m$ the following equation has a solution: $$ \begin{pmatrix} 0&0&1\\ 1&0&1\\ 1&0&2\\ 1&1&0 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix}= \begin{pmatrix} 1\\2\\m\\5 \end{pmatrix}\tag{1} $$ Working on column vectors might make some observation easier. Note that (1) is equivalent to $$ x\begin{pmatrix} 0\\1\\1\\1 \end{pmatrix} +y\begin{pmatrix} 0\\0\\0\\1 \end{pmatrix} +z\begin{pmatrix} 1\\1\\2\\0 \end{pmatrix}=\begin{pmatrix} 1\\2\\m\\5 \end{pmatrix}\tag{2} $$ It is very easy to observe that when $m=3$, (2) has a solution: $x=z=1$, $y=4$. This rules out A,B,C. To see $m=3$ is necessarily true, note that (2) implies: $$ 0x+0y+1z=1\\ 1x+0y+1z=2\\ 1x+0y+2z=m $$ To make the linear system consistent so that it has a solution, one must have $1+2=m$ by observing that the coefficients of the first two rows add up to those of the third row.