Question

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

4.$\mathit{A}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{2}& \mathbf{3}\\ \mathbf{6}& \mathbf{9}\end{array}\mathbf{\right]}$

Short answer

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

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 $A$:

$\left[\begin{array}{ccc}2& 3& 0\\ 6& 9& 0\end{array}\right]R_2\text{}\to R_2-3R_1\left[\begin{array}{ccc}2& 3& 0\\ 0& 0& 0\end{array}\right]R_1\to \frac{1}{2}{R}_1\left[\begin{array}{ccc}1& \frac{3}{2}& 0\\ 0& 0& 0\end{array}\right]$

The above equation gives $x_1+\frac{3}{2}{x}_2=0$. This implies$x_1=-\frac{3}{2}{x}_2$ .

From the above calculation, we can say that the solution set of the linear system is $\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}-\frac{3}{2}{x}_2\\ x_2\end{array}\right]$. So, we have$\overrightarrow{x}=x_2\left[\begin{array}{c}-\frac{3}{2}\\ 1\end{array}\right]$ .

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