Math 220 Quiz Solution

The Quiz

Find the kernel of the linear map $${\mathbf{R}}^3\stackrel{f}{\to }{\mathbf{R}}^2$$ that is defined for $$x=\left(\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right)\text{in}{\mathbf{R}}^3$$ by $$f\left(x\right)=\left(\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 0\\ \hfill 2& \hfill 1& \hfill 5\end{array}\right)\left(\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\end{array}\right)\text{.}$$

Recommended Solution

1. The kernel of $f$ is the set of all $x$ in ${\mathbf{R}}^3$ such that $f\left(x\right)=\stackrel{\rightharpoonup }{0}$.

2. Finding this kernel amounts to solving a system of 2 linear equations in 3 variables.

3. Perform elementary row operations on the given matrix so as to maneuver it into reduced row echelon form. (Augmenting it by a zero column would be a waste of time since a zero column remains a zero column under any row operation.) $$\left(R_2\to R_2-2R_1\right)\left(\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill 5& \hfill 5\end{array}\right)$$ $$\left(R_2\to \left(1⁄5\right)R_2\right)\left(\begin{array}{ccc}\hfill 1& \hfill -2& \hfill 0\\ \hfill 0& \hfill 1& \hfill 1\end{array}\right)$$ $$\left(R_1\to R_1+2R_2\right)\left(\begin{array}{ccc}\hfill 1& \hfill 0& \hfill 2\\ \hfill 0& \hfill 1& \hfill 1\end{array}\right)$$

4. The resulting system of linear equations is: $$\left\{\begin{array}{ccc}\hfill x_1+2x_3& \hfill =\hfill & 0\hfill \\ \hfill x_2+x_3& \hfill =\hfill & 0\hfill \end{array}\right.$$

   $$\left\{\begin{array}{ccc}\hfill x_1& \hfill =\hfill & -2x_3\hfill \\ \hfill x_2& \hfill =\hfill & -x_3\hfill \end{array}\right.$$

5. Conclusions:

   • The kernel has the parametric form $$x=t\left(\begin{array}{c}\hfill -2\\ \hfill -1\\ \hfill 1\end{array}\right)\text{.}$$

   • The kernel is the linear span (in ${\mathbf{R}}^3$) of $\left(-2,-1,1\right)$.