How do you solve Ax=B given A=((2, 0, 0), (-1, 2, 0), (-2, 4, 1)) and B=((4), (10), (11))?

Jul 30, 2016

Column vector $x = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right] = \left[\begin{matrix}2 \\ 6 \\ 50\end{matrix}\right]$

Explanation:

You may regain the original system of linear equations represented in the given matrix form by applying the definition of matrix multiplication using the Euclidean inner product in ${\mathbb{R}}^{3}$, together with the definition of matrix equality to obtain :

$2 x + 0 y + 0 z = 4$ .......1.
$- x + 2 y + 0 z = 10$ .......2.
$- 2 x + 4 y + z = 11$ ........3.

where $x$ is the column vector $x = \left[\begin{matrix}x \\ y \\ z\end{matrix}\right]$.

Solving equation 1 we get $x = 2$.

Substituting this into equation 2, we get $y = 6$.

Substituting this into equation3 we get $z = 50$.