I have a Matrix:
$$A= \pmatrix{1 & -2 & 1 \\ -1 & 3 & 2 \\ 0 & 1 & 4}$$
My task is to find $X$ from:
$$A * X = \pmatrix{4 & 0 & -3 & 1 \\ 1 & 5 & 2 & -1 \\ 0 & 1 & -1 & 2}$$
My problem is, that i dont know how to do this. I mean i could build several equations like:
$$1 * x1,1 -2 *x1,2 + 0* x1,3 = 4$$
But i think this would take to much time! So what could i do?
The solution for $X$ should be:
$$X = \pmatrix{49 & 38 & -5 & -13 \\ 20 & 17 & -1 & -6 \\ -5 & -4 & 0 & 2}$$
If you denote by $x_1$, $x_2$, $x_3$ and $x_4$ the columns of $X$ and by $b_1$, $b_2$, $b_3$ and $b_4$ the columns of the given $3\times 4$ matrix (call it $B$), you have essentially to solve the linear systems $$ Ax_i=b_i\qquad(i=1,2,3,4) $$ If you consider the “multiaugmented” matrix $$ \left[\begin{array}{c|c|c|c|c}A&b_1&b_2&b_3&b_4\end{array}\right] $$ you can solve them all at once, by reducing the matrix to echelon form. If you find a pivot in the last four columns, your problem has no solution. If the rank of $A$ is $3$ the problem has exactly one solution. If the rank of $A$ is $1$ or $2$, and no pivot is found in the last four columns, the problem has infinitely many solutions.
