rref matrix equations k2 - 5 This question is about reduced row echelon form, Gauss-Jordan, inverting matrices, and solving systems of equations.
I try to solve a system of equations with matrices. I know what operations are allowed, but I just seem to arrive at the wrong conclusion 50 % of the times. So here are three problems, each with my calculation. My hope is to clarify if I:


*

*am making a careless misstake, and where those mistakes are (if so, I may have to do these problems in a slower pace) 

*do not know the theory well enough (don't make the correct steps)

*use a bad or "not smart" way of attacking the problem. (for example, if I do row1 + row2 when I shoul have taken row1 - row3).


 A: Note that your second attept is correct, almost (you simply did not row reduce completely): You want to multiply through the first two rows by $(-1)$. Then swap Row 1 and R 2. Then simply find a common denominator for your last common: You'll have the matrix:
$$\begin{pmatrix} 1 & 0 & 0 &| &\frac 6{31} \\ 0 & 1 & 0 &| & \frac{55}{31} \\ 0 & 0 & 1 &|& -\frac{22}{31}\end{pmatrix}$$

A few suggestions (if you were starting from scratch):
Why not just switch rows, swapping rows $1$ and $3$? That will greatly simplify the row reduction!
$$\begin{pmatrix} 10 & 1& 1&| & 3\\ 3 & 2 & 3 &| & 2\\ 2 & 3 & 1 &| & 5\end{pmatrix} \to \begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 10 & 1 & 1 & | & 3\end{pmatrix}$$
Now take $(-5R_1 + R_3) \to R_3, \quad \cdots$
$$\begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 10 & 1 & 1 & | & 3\end{pmatrix} \to
\begin{pmatrix} 2 & 3 & 1 &| & 5 \\ 3 & 2 & 3 &| &2\\ 0 & -14 & -4 &| & -12\end{pmatrix}$$
Now multiply the bottom row by $\left(-\frac 12\right).\;$ Then take $\left(-\frac 32 R_1 + R_2\right) \to R_2$.
See what more you can do, then.
