Find the Vector Form Solution to the Matrix Equation $A\mathbf{x}=\mathbf{0}$

Problem 701

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$, where $A=\begin{bmatrix}
1 & 1 & 1 & 1 &2 \\
1 & 2 & 4 & 0 & 5 \\
3 & 2 & 0 & 5 & 2 \\
\end{bmatrix}$. Also, find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$.

 
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Solution.

Find the vector form solution $\mathbf{x}$ of the equation $A\mathbf{x}=\mathbf{0}$

We reduce the augmented matrix as follows:
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
1 & 2 & 4 & 0 & 5 & 0 \\
3 & 2 & 0 & 5 & 2 & 0 \\
\end{array} \right] \xrightarrow[R_3-3R_1]{R_2-R_1}
\left[\begin{array}{rrrrr|r}
1 & 1 & 1 & 1 &2 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & -1 & -3 & 2 & -4 & 0 \\
\end{array} \right] \\[6pt] \xrightarrow[R_3+R_2]{R_1-R_2}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 2 &-1 & 0 \\
0 & 1 & 3 & -1 & 3 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right] \xrightarrow[R_2+R_3]{R_1-2R_3}
\left[\begin{array}{rrrrr|r}
1 & 0 & -2 & 0 &1 & 0 \\
0 & 1 & 3 & 0 & 2 & 0 \\
0 & 0 & 0 & 1 & -1 & 0 \\
\end{array} \right].
\end{align*}
Hence, the solution of the system is
\begin{align*}
x_1&=2x_3-x_5\\
x_2&=-3x_3-2x_5\\
x_4&=x_5,
\end{align*}
where $x_3, x_5$ are free variables.

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The vector form of the general solution is
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
2x_3-x_5 \\
-3x_3-2x_5 \\
x_3 \\
x_5 \\
x_5
\end{bmatrix}\\[6pt] &=x_3\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}+x_5\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}.
\end{align*}

Find two linearly independent vectors $\mathbf{x}$ satisfying $A\mathbf{x}=\mathbf{0}$

For example, setting $x_3=1, x_5=0$, we see that $\begin{bmatrix}
2 \\
-3 \\
1 \\
0 \\
0
\end{bmatrix}$ is a solution. Similarly, setting $x_3=0, x_5=1$, we see that $\begin{bmatrix}
-1 \\
-2 \\
0 \\
1 \\
1
\end{bmatrix}$ is another solution.
It is straightforward to check that these two vectors are linearly independent.

Common Mistake

This is a midterm exam problem of Lienar Algebra at the Ohio State University.

For the second part, some students chose the zero vector.
But note that the zero vector and another nonzero vector are always linearly dependent.

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