Solution of equation system Ax=12x

$$1=x_1+x_2+x_3=\dfrac{3}{2}x_3\color{red}+\dfrac{3}{2}x_3+x_3=\color{red}4x_3 \implies x_3=\color{red}{\dfrac{1}{4}}.$$

So the only solution satisfying the constraints is $\Big[\dfrac{3}{2},\dfrac{3}{2},\dfrac{1}{4}\Big].$

$\color{red}{\text{Edited}}$ thanks to @amd and @J.W.Tanner.


It should be $rref(A)$=$\begin{bmatrix}1 & 0 & -3/2\\0 & 1 & \color{red}-3/2\\0 & 0 & 0\\\end{bmatrix}$.

So $x_1=x_2=\frac32x_3$.

To get $x_1+x_2+x_3=\frac32x_3+\frac32x_3+x_3=1$, take $x_3=\frac14$.

Tags:

Linear Algebra

Matrices

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