Find value of parameter so that three vectors are linearly dependent - I get three equations with four variables

Question

Thanks for taking your time.

I have come across an exercise that says.

Given three vectors

$ \vec{m} = (3t-1)\vec{i} +2\vec{j} +2\vec{k} $

$ \vec{n} = (3-8t)\vec{i} +(t-2)\vec{j} -3t\vec{k} $

$ \vec{p} = (2t-6)\vec{i} +4\vec{j} -3\vec{k} $

Find the value of parameter $t$ so that the three vectors are linearly dependent.

So I do know that I need to find three scalars, all of which are non-zero value. Written:

$ \alpha \vec{m} + \beta \vec{n} + \gamma \vec{p} = \vec{0} $

We now form the three equations:

$3\alpha t- \alpha +3\beta -8\beta t +2\gamma t -6\gamma = 0$

$2\alpha +\beta t - 2\beta +4\gamma =0 $

$2\alpha - 3\beta t - 3\gamma = 0$

Now, if the trivial solution to this is of course $\alpha = 0, \beta =0, \gamma =0$ but I need to find parameter $t$ so that the tree vectors are linearly dependent.

However, how do I do that because what I have got here is a system of three equations with four variables!

Can anyone help? Thanks!

Answer 1

hint

It is so easier to use the fact that

$\vec{m} \; , \; \vec{n} \;\text{ and } \;\vec{p}$

are linearly dependent if and only if

$$det(\vec{m},\vec{n},\vec{p})=0$$

$$=-3(3t-1)(t-2)-2\Bigl(3t(3t-6)-3(3-8t)\Bigr)+2\Bigl(4(3-8t)-(t-2)(2t-6)\Bigr)$$

and solve for $t$.

Answer 2

The determinant of a matrix vanishes iff its rows/columns are linearly dependent, so a straightforward way to solve this problem is to solve the equation $\det(\vec m,\vec n,\vec p)=0$, as hamam_Abdallah explains in his answer. However, we can proceed with your approach as well.

For the system of homogeneous equations that you’ve set up to have a nontrivial solution, its coefficient matrix must be rank-deficient. (This is equivalent to the vanishing determinant above.) So, proceed as you might normally do to solve a system of linear equations: form the coefficient matrix and row-reduce. We start with $$\begin{bmatrix}3t-1&2&2\\3-8t&t-2&-3\\2t-6&4&-3\end{bmatrix}.$$ The parameter $t$ is in some inconvenient places, so we’ll swap the first and third columns (this doesn’t affect the rank of the matrix) and while we’re at it, swap the second and third rows to get that $t$ off the main diagonal: $$\begin{bmatrix}2&2&3t-1\\-3&4&2t-6\\-3&t-2&3-8t\end{bmatrix}.$$ After a few elementary row operations, we have the upper-triangular matrix $$\begin{bmatrix}2&2&3t-1\\0&7&\frac12(13t-15)\\0&0&-\frac1{14}(13t^2+47t-36)\end{bmatrix}.$$ We want the last row to consist of zeros, which gives you a quadratic equation in $t$ to solve. In fact, the lower-right element of this last matrix is a multiple of the determinant of the original coefficient matrix.