Question

For each matrix$\mathit{A}$ in exercises 1 through 13, find vectors that span the kernel of $\mathit{A}$. Use paper and pencil.

2.$\mathit{A}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{2}& \mathbf{3}\end{array}\mathbf{\right]}$

Short answer

The kernel of$A$ is $\ker \left(A\right)=span(\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right])$.

Step-by-step solution

The kernel of a matrix

The kernel of a matrix $\mathit{A}$is the solution set of the linear system $\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\mathbf{0}$.

Find kernel of the given matrix

Solve the linear system$A\overrightarrow{x}=0$by reduced row-echelon form of$A$:

$\left[\begin{array}{cccc}1& 2& 3& 0\end{array}\right]$

The above equation gives$x_1+2x_2+3x_3=0$. This implies$x_1=-2x_2-3x_3$.

From the above calculation, we can say that the solution set of the linear system is$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-2x_2-3x_3\\ x_2\\ x_3\end{array}\right]$. So, we have$\overrightarrow{x}=x_2\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right]+x_3\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right]$.

Thus,$\ker \left(A\right)=span(\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right])$$\ker \left(A\right)=span(\left[\begin{array}{c}-2\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ 0\\ 1\end{array}\right])$.