Matrix equation $Ax=b$ with variable

Question

"Let a be an arbitrary real number. A matrix A and a vector b is given by $$\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & a-1 & a & 0 \\ 0 & a & a-1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{ and } \begin{bmatrix} 1 \\ 1 \\ -1 \\ -1 \end{bmatrix}$$

Find the value of a for which the matrix equation $A \cdot x = b$ has the the solution $$x= \begin{bmatrix} 2 \\ -2 \\ 0 \\ -1 \end{bmatrix} + t \cdot \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}, t \in \mathbb{R}$$ "

I am not really sure how this can be approached.

Answer 1

Guide:

One possible way is to let $t=0$ to obtain a particular solution $x_0$.

Write out the matrix product $Ax_0$ and equate it to $b$.

You can then solve for $a$ since we will get linear equation in $a$.

After which verify that we can obtain the solution.

Alternative method to solve for $a$ is to note that $[0,1,1,0]^T$ is in the nullspace, again, you will get a linear equation in $a$.

Answer 2

To satisfy the given condition it,s necessary that the vector multiplied by $t$ belong to the Null Space or Ker(A), thus

$$A \ \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$

that gives you the correct value for a.

Finally check that the given solution work for the value of $a$ that you find.