Solve the system for $x_1, x_2, x_3 $ when $\lambda = 1$

Question

$$\begin{array}{rcrcrcl} 2x_1 & - & x_2 & & & = & \lambda x_1\\ 2x_1 & - & x_2 & + & x_3 & = & \lambda x_2\\ -2x_1 & + & 2x_2 & + & x_3 & = & \lambda x_3\end{array}$$

So when $\lambda = 1$ we have

$$\begin{array}{1} \,\,\,\,2x_1 - \,\,x_2 \quad\quad\,\, = x_1\\ \,\,\,\, 2x_1 - \,\,x_2 + x_3 \,= x_2\\ -2x_1 + 2x_2 + x_3 = x_3 \end{array}$$

So then I brought over the right-hand side to the left-handside.

$$\begin{array}{1} \,\,\,\,\,\,x_1 - \,\,x_2 \quad\quad\,\,\,\, = 0\\ \,\,\,\, 2x_1 - 2x_2 + x_3 \,= 0\\ -2x_1 + 2x_2 + \quad\,\,\, = 0 \end{array}$$

So now I reduced it go get the following augmented matrix:

$$\begin{bmatrix} 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

From here I'm stuck because the answer in the back of the book says that
$x_1 = x_2 = -\frac12 s$, $x_3 = s$

I thought it would have been more like:
$x_1 = x_2 = s$, $x_3 = 0$

What am I doing wrong?

Answer 1

Your solution is correct, the books is not.

If you plug in your solution to the original equation, you get

$2s-s=s\\ 2s-s+0=s\\ -2s+2s+0=0$

and all three equations are correct.

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On the other hand, if you plug in the book's solution, the first equation becomes

$$2\cdot(-\frac12)s -(-\frac12 s) = s$$ which, already, is clearly not true since it is equivalent to $$\frac32 s = s$$ which is only ever true for $s=0$.

Answer 2

Obviously, the solution from the back of your book is not correct. Take $s=2$. From the solution from the back of your book, you get that a solution of the system (when $\lambda=1$) is $x_1=x_2=-1$ and $x_3=2$. But then $2x_1-x_2+x_3=1\neq2=x_3$.