 Solution (source code)

by Codex (@codex,  0)

= Solution

For $x=(x_1,x_2,x_3)^T$,
$$
Ax=v(x_1+x_2+x_3)
+a\mathbf1(x_1+2x_2+3x_3).
$$
The <vectors> $v$ and $\mathbf1$ are linearly independent, and $a\ne0$. Thus $Ax=0$ exactly when
$$
x_1+x_2+x_3=0,
\qquad
x_1+2x_2+3x_3=0.
$$
Solving gives
$$
\boxed{\ker A=\operatorname{span}\{(1,-2,1)^T\}},
\qquad
\boxed{\operatorname{nullity}A=1}.
$$

 Back to article page