Question

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

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

Short answer

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

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 $A$:

$\left[\begin{array}{cccc}1& 2& 3& 0\\ 1& 3& 2& 0\\ 3& 1& 1& 0\end{array}\right]\begin{array}{c}{R}_2\to R_2-R_1\\ R_3\to R_3-3R_1\end{array}\left[\begin{array}{cccc}1& 2& 3& 0\\ 0& 1& -1& 0\\ 0& -4& -8& 0\end{array}\right]\begin{array}{c}{R}_1\to R_1-2R_2\\ R_3\to R_3+4R_2\end{array}\left[\begin{array}{cccc}1& 0& 5& 0\\ 0& 1& -1& 0\\ 0& 0& -12& 0\end{array}\right]$

Further solve as:

$\left[\begin{array}{cccc}1& 0& 5& 0\\ 0& 1& -1& 0\\ 0& 0& -12& 0\end{array}\right]R_3\to -\frac{1}{12}{R}_3\left[\begin{array}{cccc}1& 0& 5& 0\\ 0& 1& -1& 0\\ 0& 0& 1& 0\end{array}\right]\begin{array}{c}{R}_1\to R_1-5R_3\\ R_2\to R_2+R_3\end{array}\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right]$

The above equation gives$x_1=0,x_2=0,x_3=0$ .

From the above calculation, we can say that the solution set of the linear system is .

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

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