Science:Math Exam Resources/Courses/MATH307/December 2012/Question 01 (d)/Solution 1

Since $\overrightarrow{a}=s_1\overrightarrow{a_1}+s_2\overrightarrow{a_2}=s_1\left[\begin{array}{c}1\\ -3\\ 0\\ 0\end{array}\right]+s_2\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]$$=\left[\begin{array}{cc}1& 0\\ -3& 0\\ 0& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}{s}_1\\ s_2\end{array}\right]$

Using the equation from part (c), we have

$B\overrightarrow{a}=\overrightarrow{b}$

$\left[\begin{array}{cccc}8& 4& 2& 1\\ 1& 1& 1& 1\\ 0& 0& 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -3& 0\\ 0& 0\\ 0& 1\end{array}\right]\overrightarrow{s}=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$

Therefore,

$C=\left[\begin{array}{cc}-4& 1\\ -2& 1\\ 0& 1\end{array}\right]$ and $\overrightarrow{c}=\overrightarrow{b}=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$.

Notice that for there to be a solution we need $\overrightarrow{c}$ in the range of C, R(C). We know that R(C) is the orthogonal complement to the nullspace of $C^T$, $N(C^T)$. Therefore, onsider the basis vector of $N(C^T)$ $\left[\begin{array}{ccc}-4& -2& 0\\ 1& 1& 1\end{array}\right]\overrightarrow{x}=0\to \left[\begin{array}{ccc}2& 1& 0\\ 0& 1& 2\end{array}\right]\overrightarrow{x}=0$

$\overrightarrow{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]x_3$

Since $\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]\cdot \left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]=2-4+1\ne 0$

In other words, $N(C^T)$ is not orthogonal to $\overrightarrow{c}$. This means that $\overrightarrow{c}$ cannot possibly be in the range of C and so there will be no solution to the problem.