How to calculate a general solution to a linear system from a kernel.

Question

My problem is a two part problem where you calculate the ker(A) then using the kernel, you are supposed to find the general solution to a linear equation. Here are the components:

Matrix A=$$ \begin{bmatrix} 1 & 2 & 3 & 2 & 1\\ 0 & 2 & 1 & 1 & 2\\ 1 & 4 & 4 & 3 & 3\\ 0 & 2 & 1 & 1 & 2\\ \end{bmatrix} $$ The linear system: $$ \begin{bmatrix} 1 & 2 & 3 & 2 & 1\\ 0 & 2 & 1 & 1 & 2\\ 1 & 4 & 4 & 3 & 3\\ 0 & 2 & 1 & 1 & 2\\ \end{bmatrix} \begin{bmatrix}X_1\\ X_2\\ X_3 \\ X_4 \\ X_5 \\ \end{bmatrix} = \begin{bmatrix} 11\\ 7 \\ 18 \\ 7 \\ \end{bmatrix}$$ I know how to solve this using an augmented matrix and I put A into RREF and got the Kernel but just don't understand how to solve it using the kernel. $$Ker(A)=X_3\begin{bmatrix} -2 \\ -.5 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} + X_4 \begin{bmatrix} -1 \\ -.5 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} + X_5 \begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}$$ If someone could explain how to do this it would be very helpful! Thank You.

Answer 1

You don't have to find anything, what you have found is a basis of \ker(A).

When you solve the system for a particular solution say $Ax_P=b$ you can add to $x_P$ a linear combination of the basis vectors of $\ker(A)$ to obtain a general solution, that is

$$x_G=x_P+av_1+bv_2+cv_3$$

since $Av_i=0$ for $v_i\in\ker(A)$.

Answer 2

If you know a single particular solution $x_p$, then the general solution of $Ax=b$ is $x_p+\ker A$. This happens because if $x_p$ and $y_p$ are two solutions, then $$ A(x_p-y_p)=Ax_p-Ay_p=b-b=0, $$ so $x_p-y_p\in \ker A$.

In this case, you check by inspection that $$ x_p=\begin{bmatrix} 1\\1\\1\\2\\1\end{bmatrix} $$ is a solution. Then the general solution is $$ \begin{bmatrix} 1\\1\\1\\2\\1\end{bmatrix}+r\begin{bmatrix} -2 \\ -1/2 \\ 1 \\ 0 \\ 0 \\ \end{bmatrix} + s \begin{bmatrix} -1 \\ -1/2 \\ 0 \\ 1 \\ 0 \\ \end{bmatrix} + t \begin{bmatrix} 1 \\ -1 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}, $$ for arbitrary choices of $r,s,t$.