Can I get this Gaussian elimination solution correct?

Question

I have "Linear Algebra an introduction" ed 2 by A.O.Morris open in front of me at Exercise 1.2 (vi):

"Find the reduced echelon matrix of:

$\begin {pmatrix} 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\ \sqrt 2 & -3 & 1 + \sqrt 2 & -1 - 2 \sqrt 2 \\ -1 & \sqrt 2 & -1 & 1 \\ \sqrt 2 - 2 & -2 + 4 \sqrt 2 & -2 - \sqrt 2 & 3 + \sqrt 2 \\ \end {pmatrix}$

The solution in the back of the book says:

$\begin {pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end {pmatrix}$

but after $10$ or so elementary row operations I get as far as this:

$\begin {pmatrix} 1 & 0 & 1 - \sqrt 2 & 1 + \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & -1 + 2 \sqrt 2 \\ \end {pmatrix}$

... and as you see, one of my rows has vanished.

I've tried this a number of ways, but it seems to boil down to the same thing.

Here is an example sequence:

In the following, $\left\langle {e_n}\right\rangle_{n \mathop \ge 1}$ denotes the sequence of elementary row operations that are to be applied to $\mathbf A$.

The matrix that results from having applied $e_1$ to $e_k$ in order is denoted $\mathbf A_k$.

$e_1 := r_3 \to r_3 + r_1$

$e_2 := r_4 \to r_4 - r_2$

$e_3 := r_4 \to r_4 + 2 r_1$

$\mathbf A_3 = \begin {pmatrix} 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\ \sqrt 2 & -3 & 1 + \sqrt 2 & -1 - 2 \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\ \end {pmatrix}$

$e_4 := r_2 \to r_2 - \sqrt 2 r_1$

$\mathbf A_4 = \begin {pmatrix} 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\ 0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\ \end {pmatrix}$

$e_5 := r_2 \leftrightarrow r_3$

$\mathbf A_5 = \begin {pmatrix} 1 & 1 - \sqrt 2 & 0 & \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\ 0 & 3 + 2 \sqrt 2 & -3 - 2 \sqrt 2 & 4 + 5 \sqrt 2 \\ \end {pmatrix}$

$e_6 := r_1 \to r_1 - r_3$

$e_7 := r_4 \to r_4 + 2 r_3$

$\mathbf A_7 = \begin {pmatrix} 1 & 2 & -1 - \sqrt 2 & 3 + 3 \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & -1 - \sqrt 2 & 1 + \sqrt 2 & -3 - 2 \sqrt 2 \\ 0 & -1 & -1 & -2 + \sqrt 2 \\ \end {pmatrix}$

$e_8 := r_1 \to r_1 - 2 r_2$

$e_9 := r_3 \to r_3 - (-1 - \sqrt 2) r_2$

$e_{10} := r_4 \to r_4 + r_2$

$\mathbf A_{10} = \begin {pmatrix} 1 & 0 & 1 - \sqrt 2 & 1 + \sqrt 2 \\ 0 & 1 & -1 & 1 + \sqrt 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & -1 + 2 \sqrt 2 \\ \end {pmatrix}$

Yes there are possibly easier ways of getting to that zero row, but this is the way I did it.

The question is: is there a mistake in the book or my algebra / arithmetic?

Answer 1

Using row reduction $A\sim \begin{bmatrix} 1 & 0 & 1-\sqrt{2} & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{bmatrix}$ so $\det A=0$ so the answer in the book is wrong.

About the row operations:

1. $R_{2} \leftrightarrow \sqrt{2}R_{1}-R_{2}$.
2. $R_{3}\leftrightarrow R_{1}+R_{3}$.
3. $ R_{4} \leftrightarrow (-\sqrt{2}+2)R_{1}+R_{4}$.
4. $R_{3}\leftrightarrow \frac{1}{1+\sqrt{2}}R_{2}+R_{3}$.
5. $ R_{4}\leftrightarrow \frac{2+\sqrt{2}}{1+\sqrt{2}}R_{2}+R_{4}$

Since the determinant of upper triangle matrix is the product of the terms on principal diagonal so $\det A= (1)(-1-\sqrt{2})(0)(-3)=0$.

Answer 2

A quick check shows that the solution provided in your textbook must definitely be wrong: The matrix is singular, i.e., its determinant is equal to $0$. Hence, the reduction in the book cannot be correct, as the result would be a regular matrix.

I have to admit that I have not yet checked every line of your arithmetic, but any correctly carried out Gaussian elimination of a singular matrix must yield at least row of zeroes. Hence, you seem to be on track here!