questions about systems of equations using matrices and row echleon

Question

I have the following matrix: $$ \left[ \begin{array}{cc|c} -1&-2&{\sqrt 2}\\ -8&2&{\sqrt 3} \end{array} \right] $$

So the first thing I do is multiply R1 by - 1 to get

$$ \left[ \begin{array}{cc|c} 1&2&-{\sqrt 2}\\ -8&2&{\sqrt 3} \end{array} \right] $$

I then believe I can multiply r1 by 8 and add it to r2. This works but why does it work? I don't understand the logic of why this balances both sides of the new equation as r1 is unchanged after this modification.

After I have done that I get:

$$ \left[ \begin{array}{cc|c} 1&2&-{\sqrt 2}\\ 0&18&{{-8\sqrt2} + {\sqrt 3}} \end{array} \right] $$

I then muliply r2 by ${1\over18}$ to get

$$ \left[ \begin{array}{cc|c} 1&2&-{\sqrt 2}\\ 0&1&{{-8\sqrt2 + \sqrt3}\over 18} \end{array} \right] $$

So from this I can say that ${y = {{-8\sqrt2 + \sqrt3}\over 18}}$

I then use the substitution method to work out x which would mean:

${x + 2.({{-8\sqrt8 + \sqrt2}\over 18})} = \sqrt2$

Could somebody explain how I can simplify this further?

Answer 1

I believe you have a typo in the second row, second column of the second matrix. It should be $2$ instead of $-2$.

The reason why you can do this is because if a solution $(x,y)$ satisfies the original two equations, it should also satisfy an equation that is obtained by multiplying one of them by a nonzero constant. Also, it should satisfy the addition of the two equations. So you can replace one of the original equations by this new form, which makes the system easier to solve.

The only simplification you can do at the end is

$$y=-\frac{4}{9}\sqrt{2}+\frac{1}{18}\sqrt{3}\\ x=-\frac{1}{9}\sqrt{2}-\frac{1}{9}\sqrt{3}$$

Answer 2

You are solving a linear system and the method that you are using is based on the fact that the solution of the system does not change if one of the equation is substituted by a linear combination of the starting equations. This justify your first steps, in which you have substitute the first equation (row $r_1$ ) by $-1\times r_1$ and then the second equation with the linear combination $8\times(-1) \times r_1 +r_2$.

For the second question: I suppose that you have a typo in your first equation (as noted in my comment) so that your solution ${y = {{-8\sqrt2 + \sqrt3}\over 18}}$ is correct, if this is correct you find $$ x= -\sqrt{2}-2\dfrac{-8\sqrt2 + \sqrt3}{18} $$ (note that there is some mistake in your post) that you can semplify a little.