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Let $a_1 = (2,1,-4,-1)$, $a_2 = (-1, -1, 3, -1)$, $a_3 = (2, -3, 4, \lambda - 14)$ and $v = (2, 2, \mu - 5, \mu + 3)$ be vectors in $\mathbb{Q}^4$ over the field of rational numbers $\mathbb{Q}$. For which values of $\lambda$ and $\mu$ can the vector $v$ be represented in more than one way as a linear combination of $a_1$, $a_2$ and $a_3$. Find two of these representations.


This means that we have to find $x_1$, $x_2$, $x_3$ s.t. $v = x_1a_1 + x_2a_2 + x_3a_3$.

I started by putting the vectors as columns in a matrix and tried to reduce it. Here is what I got: $\begin{pmatrix}2 & -1 & 2 & 2 \\ 1 & -1 & -3 & 2 \\ -4 & 3 & 4 & \mu-5 \\ -1 & -1 & \lambda - 14 & \mu + 3\end{pmatrix} \rightarrow \begin{pmatrix}1 & 0 & 5 & 0 \\ 0 & 1 & 8 & -2 \\ 0 & 0 & \lambda - 1 & \mu + 1 \\ 0 & 0 & 0 & \mu + 1\end{pmatrix}$

If $\mu \neq -1$ the system of equations is incompatible and therefore there is no solution. If $\mu = -1$ the matrix simplifies further to: $\begin{pmatrix}1 & 0 & 5 & 0 \\ 0 & 1 & 8 & -2 \\ 0 & 0 & \lambda - 1 & 0\end{pmatrix}$. If $\lambda \neq 1$ it follows that $x_1 = 0$, $x_2 = -2$ and $x_3 = 0$. Therefore, assuming $\lambda \neq 1$ then $(0, -2, 0)$ is a solution to the system. Otherwise the system simplifies to $\begin{pmatrix}1 & 0 & 5 & 0 \\ 0 & 1 & 8 & -2\end{pmatrix}$. This means that $x_1 + 5x_3 = 0 \iff x_1 = -5x_3$ and $x_2 + 8x_3 = -2 \iff x_2 = -2 - 8x_3$. Let $p = x_3$, then the solutions of the system are of the form: $(-5p, -2 - 8p, p)$, $\forall p\in\mathbb{Q}$ assuming $\mu = -1$ and $\lambda = 1$. Two possible representations can be obtained by taking $p = 0$ which makes $v = -2a_2 = (2, 2, -6, 2)$ and taking $p = -1$ which makes $v = 5a_1 + 6a_2 - a_3 = \dots = (2, 2, -6, 2)$. I'd be very grateful if someone could tell me if my way of solving this problem is correct and point out any potential mistakes I've made. Thank you!

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    $\begingroup$ Why do you say $\lambda\neq 1$ yields no solution? (Apart from this case, assuming you've made no numerical errors in the first reduction, your solution is correct and quite well written :) ) $\endgroup$
    – Al.G.
    Nov 21, 2023 at 21:23

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As Al.G. have pointed out, you missed the solution only for $\lambda\neq1$.

$$ \begin{cases} L_1:2x_1-x_2+2x_3 = 2\\ L_2:x_1-x_2-3x_3 = 2\\ L_3:-4x_1+3x_2+4x_3 = \mu-5\\ L_4:-x_1-x_2+x_3(\lambda-14) = \mu+3 \end{cases}$$ $$\begin{cases} L_1:2x_1-x_2+2x_3 = 2\\ L_2^\prime=L_1-2L_2:x_2+8x_3 = -2\\ L_3^\prime=2L_1+L_3:x_2+8x_3 = \mu-1\\ L_4^\prime=L_1+2L_4:-3x_2+2x_3(\lambda-13) =2(\mu+4) \end{cases} $$ $$\begin{cases} L_1:2x_1-x_2+2x_3 = 2\\ L_2^\prime:x_2+8x_3 = -2\\ L_3^{\prime\prime}=-L_2^\prime+L_3^\prime:0= \mu+1\\ L_4^{\prime\prime}=3L_2^\prime+L_4^\prime:2x_3(\lambda-1) =2(\mu+1) \end{cases} $$ From $L_3^{\prime\prime}$ we conclude that $$\boxed{\mu=-1}$$ Substituting this into $L_4^{\prime\prime}$, we obtain $$2x_3(\lambda-1)=0.$$ Case 1: If $\boxed{\lambda\neq1}$, then the solution is $$\boxed{x_1=0 \ ; \ x_2=-2 \ ; \ x_3=0}.$$ Case 2: If $\boxed{\lambda=1}$, then the solution is $$\boxed{x_1=-5x_3 \ ; \ x_2=-2-8x_3}.$$

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