Solve the system

Question

Solve the system: The last column is the vector b $$ \begin{bmatrix} 1 & 1 & 4 & -5 \\ 4 & 3 & -5 & 8 \\ \end{bmatrix} $$

I reduced it down to

$$ \begin{bmatrix} 1 & 0 & -17 & 23 \\ 0 & 1 & 21 & -28 \\ \end{bmatrix} $$

Now I have to express it in terms of:

$$ \begin{bmatrix} x1 \\ x2 \\ x3 \\ \end{bmatrix} $$

x3 is free and I believe the answer to be something along the lines of

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

Not sure what the ? values are.

Answer 1

Let every column correspond to a $x_n$. So for the first row, the equation will be something like $$(1)x_1 + (0)x_2 + (-17)x_3 = 23$$ and the second row equation will be $$(0)x_1 + (1)x_2 + (21)x_3 = -28$$ then you just solve for $x_1$ and $x_2$ where $x_3$ is a free variable.

Answer 2

$x + y + 4z = -5 \Rightarrow x = -5 - y - 4z$ (i)

$4x + 3y -5z = 8 \Rightarrow -20 -4y - 16z + 3y - 5z = 8 \Rightarrow - 21z - 28 = y(z)$

(i) $\Rightarrow x(z) = -5 +21z + 28 - 4z = 17z + 23$

$(x,y,z) = (17z + 23, - 21z - 28, z) = z (17, - 21, 1) + (23, - 28, 0)$