Find a non-zero integer matrix $X$ such that $XA=0$ where $X,A,0$ are all $4 \times 4$ Let $A$ be the following $4 \times 4$ matrix.
\begin{bmatrix}1&2&1&3\\1&3&2&4\\2&5&3&7\\1&4&1&5\end{bmatrix}
How can we find a non-zero integer matrix $C$ such that $CA = 0_{4 \times 4}$
Note that $0$ is a $4 \times 4$ matrix. 
 A: HINT
left-mutipliction by a matrix is equivalent to taking linear combinations of rows of $A$
Use the fact that
Row 2 + Row 1 = Row 3
A: Note that solving $XA=0$ is equivalent to solving $A^\top Y=0$. Since 
$$
\DeclareMathOperator{rref}{rref}\rref A^\top = 
\left[\begin{array}{rrrr}
1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]
$$
it follows that all solutions to $A^\top \vec y=\vec 0$ are of the form
$$
\vec y=\lambda
\left[\begin{array}{r}
1 \\
1 \\
-1 \\
0
\end{array}\right]
$$
Now, let $\lambda_1,\dotsc,\lambda_4$ be your four favorite numbers. Then
$$
Y=
\begin{bmatrix}
\lambda_1\vec b & \lambda_2\vec b & \lambda_3\vec b & \lambda_4\vec b 
\end{bmatrix}
$$
is a $4\times 4$ matrix that solves $A^\top Y=0$ where 
$$
\vec b=
\left[\begin{array}{r}
1 \\
1 \\
-1 \\
0
\end{array}\right]
$$
Taking $X=Y^\top$ then solves your original problem.
A: Hint : The rank of $A$ is $3$ so : $$\exists P,Q \in GL_n(\mathbb R),A=P\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q$$
$$CA=0 \Rightarrow CP\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}Q=0\Rightarrow C\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&0\end{bmatrix}=Q^{-1}P^{-1}$$
Calculate $P$ and $Q$ to find a necessary condition on $C$ and take a matrix fulfilling this condition and verify it is indeed a solution to your problem.
A: Solve the equation $A^Tx = 0$. If $A$ is not invertible then it must have a nonzero solution $\vec x_0$.
Then set $B = \begin{pmatrix} \vec x_0 \\ \vec x_0 \\ \dots \\ \vec x_0\end{pmatrix}$ -- the matrix having $\vec x_0$ as rows. Since $\vec x_0^T A = 0$, it means that $BA = 0$ and $B \neq 0$.
Since $\vec x_0$ is a vector with rational entries, we can represent it as $\vec x_0 = \begin{pmatrix} \dfrac{p_1}{q_1} &  \dfrac{p_2}{q_2} & \dots & \dfrac{p_n}{q_n}\end{pmatrix}$. Let $k = \operatorname{lcm}(q_1, q_2, \dots, q_n)$. Then $C = kB \neq 0$ is a desired matrix.
A: One option is use the projector onto the $\color{red}{null}$ space 
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A^{*}}\right)}} = \mathbf{I}_{4} - \mathbf{A}\mathbf{A}^{+}
$$
The target matrix and pseudoinverse are
$$
\mathbf{A} = 
\left(
\begin{array}{cccc}
 1 & 2 & 1 & 3 \\
 1 & 3 & 2 & 4 \\
 2 & 5 & 3 & 7 \\
 1 & 4 & 1 & 5 \\
\end{array}
\right),
\qquad
\mathbf{A}^{+}
=
\frac{1}{18}
\left(
\begin{array}{rrrr}
 26 & -19 & 7 & -9 \\
 -16 & 11 & -5 & 9 \\
 -12 & 15 & 3 & -9 \\
 10 & -8 & 2 & 0 \\
\end{array}
\right)
$$
The projector matrix is
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A}\right)}} = \mathbf{I}_{4} - \mathbf{A}^{+}\mathbf{A} = 
\frac{1}{3}
\left(
\begin{array}{rrrr}
 1 & 1 & -1 & 0 \\
 1 & 1 & -1 & 0 \\
 -1 & -1 & 1 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
Verify:
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A}\right)}} \mathbf{A}= 
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
If you look carefully at the solution matrix, you will appreciate the insight of @gt6989b.
