Math subject GRE Exam 9768 Q.16

Question

The question is in the following picture:

And the answer is (D).

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At first I agreed with this answer when I added the given three vectors component wise,but when I added the forth component in the three vectors I understood that it is impossible that we can get 5, so my opinion changed to the choice (A), so what is the misconception that I have or where is my wrong thinking, could anyone help me please?

Answer 1

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

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

Answer 2

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 $m = 3$.

Answer 3

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.

Answer 4

The thing you are overlooking is that in a linear combination you are allowed to multiply each vector by whatever scalar constant you want.

This means that getting, say, a 5 is always possible as long as at least one of the corresponding components is not zero as there are many coefficient choices that would accomplish that.

The only real caveat when checking linear dependence is that the same coefficients have to also yield the correct values for all the other components at the same time. This is where the formalism with matrices and equation systems comes in handy.