# Represent a vector using given vectors

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!

• 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 :) ) Nov 21, 2023 at 21:23

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 $$\boxed{x_1=-5x_3 \ ; \ x_2=-2-8x_3}.$$