8
$\begingroup$

The question is in the following picture:

enter image description here]

And the answer is (D).


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?

$\endgroup$
  • 1
    $\begingroup$ Remember: linear combinations aren't just adding vectors together - you can subtract, or not include one in the combination. You need $m$ so that the given vector is in the span of the other three. $\endgroup$ – Sean Roberson Jul 18 '17 at 21:33
  • 4
    $\begingroup$ $(1,2,3,5)=1(0,1,1,1)+4(0,0,0,1)+1(1,1,2,0)$ $\endgroup$ – Shuri2060 Jul 18 '17 at 21:35
  • 1
    $\begingroup$ You can get the vector by using the coefficients (1,4,1). But how do you prove that m=3 is the only possible solution? $\endgroup$ – Quantaliinuxite Jul 18 '17 at 21:36
  • 1
    $\begingroup$ @Quantaliinuxite This is usually done through gaussian elimination and showing the system is inconsistent if $m\neq 3$. (In this kind of test, you don't need to prove it, and only need to convince yourself that this is the case. Pedro Tamaroff's answer does this) $\endgroup$ – Shuri2060 Jul 18 '17 at 21:37
  • 5
    $\begingroup$ Purely out of curiosity: are you taking the math GRE because you plan to go to grad school in math, or in some other discipline which requires the math GRE? $\endgroup$ – carmichael561 Jul 18 '17 at 21:42
14
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ (This is essentially the determinant argument of user "Cauchy" put in simpler terms.) $\endgroup$ – Pedro Tamaroff Jul 19 '17 at 6:53
17
$\begingroup$

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 $m = 3$.

$\endgroup$
  • 3
    $\begingroup$ Could you please explain more on how This is equivalent to studying the invertibility of the matrix? $\endgroup$ – ajay Jul 19 '17 at 5:52
  • 1
    $\begingroup$ @ajay see item $4$ in mathworld.wolfram.com/InvertibleMatrixTheorem.html $\endgroup$ – Cauchy Jul 19 '17 at 6:47
  • $\begingroup$ I think you made a sign error when computing the determinant (you computed it's opposite). Good answer nonetheless $\endgroup$ – Astyx Jul 19 '17 at 9:29
  • $\begingroup$ @Astyx I always miss the $(-1)^{i+j}$! Fixed now, thanks. $\endgroup$ – Cauchy Jul 19 '17 at 9:58
  • $\begingroup$ It's annoying indeed ! Glad to help $\endgroup$ – Astyx Jul 19 '17 at 10:04
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.