I have a such system
$$ \left\{ \begin{array}{cl} 3x_{1} + 2x_{2} + x_{3} = -1 \\ 7x_{1} + 6x_{2} + 5x_{3} = 5 \\ 5x_{1} + 4x_{2} + 3x_{3} = 2 \end{array} \right. $$
And I try to show, show the compatibility system of linear equations.
My attempt:
- Write the matrix of coefficients of this system:
$$ A = \begin{bmatrix} 3 & 2 & 1 \\ 7 & 6 & 5 \\ 5 & 4 & 3 \end{bmatrix} $$
- After write the following adjoint system $A^{*}X=0$:
$$ A^{*}X=0 \Rightarrow \begin{bmatrix} 3 & 7 & 5 \\ 2 & 6 & 4 \\ 1 & 5 & 3 \end{bmatrix} \cdot \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$
The general solution of this system is $X = [C_{1},C_{1},-2C_{1}]^{T}$, where $C_{1}$ is arbitrary constant.
- Now, consider this theorem, if system is compatibility, then vector of free coefficients of this system $B = [-1,5,2]^{T}$ is orthogonal to all vectors $X = [C_{1},C_{1},-2C_{1}]^{T}$.
If I show, that $C_{1}= 1, \Rightarrow X = [1,1,-2]^{T}$, this vector is orthogonal, because dot product is zero:
$$ \overline{a} \cdot \overline{b} = a_x · b_x + a_y · b_y + a_z · b_z = 1 · (-1) + 1 · 5 + (-2) · 2 = - 1 + 5 - 4 = 0 $$
My troubleshooting:
How can I correctly and in detail show that the vector of free terms is orthogonal to any vector from the solution space, which is obtained due to the fact that $C_{1}$ is arbitrary constant?
Your notation $A^*$ means the transpose of matrix $A$, namely $A^*=A^T$. Your $X$ is actually representing the kernel of $A^T$. We know
$$\text{span}\{X\}=\ker (A^T)\perp \text{row} (A^T)=\text{col} (A)$$
If the equation system is compatible, it means your vector $B\in \text{col} (A)$, hence
$$B\perp \text{span}\{X\}$$