# If Two Vectors Satisfy $A\mathbf{x}=0$ then Find Another Solution

## Problem 395

Suppose that the vectors
$\mathbf{v}_1=\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix} -4 \\ 0 \\ -3 \\ -2 \\ 1 \end{bmatrix}$ are a basis vectors for the null space of a $4\times 5$ matrix $A$. Find a vector $\mathbf{x}$ such that
$\mathbf{x}\neq0, \quad \mathbf{x}\neq \mathbf{v}_1, \quad \mathbf{x}\neq \mathbf{v}_2,$ and
$A\mathbf{x}=\mathbf{0}.$

(Stanford University, Linear Algebra Exam Problem)

## Solution.

We are asked to find a vector $\mathbf{x}$ in the null space of $A$, which are not $\mathbf{0}, \mathbf{v}_1, \mathbf{v}_2$.

Recall that the null space is a vector space. Thus, any linear combination of vectors in the null space is still in the null space.

Since $\mathbf{v}_1, \mathbf{v}_2$ are basis of the null space of $A$, they are in particular vectors in the null space of $A$.
Thus, for example,
$\mathbf{x}=2\mathbf{v}_2=\begin{bmatrix} -4 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ is an element of the null space and it is not equal to $\mathbf{0}, \mathbf{v}_1, \mathbf{v}_2$.

Another example is
$\mathbf{x}=\mathbf{v}_1+\mathbf{v_2}=\begin{bmatrix} -6 \\ 1 \\ -3 \\ -2 \\ 1 \end{bmatrix}.$

In general, you can prove that any vector of the form
$\mathbf{x}=c_1\mathbf{v}_1+c_2\mathbf{v}_2,$ where $c_1, c_2$ are scalars such that $(c_1, c_2)\neq (0,0), (1,0), (0, 1)$, satisfied the required conditions.

##### For Which Choices of $x$ is the Given Matrix Invertible?
Determine the values of $x$ so that the matrix \[A=\begin{bmatrix} 1 & 1 & x \\ 1 &x &x \\...