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

Since ${\vec {a}}=s_{1}{\vec {a_{1}}}+s_{2}{\vec {a_{2}}}=s_{1}{\begin{bmatrix}1\\-3\\0\\0\end{bmatrix}}+s_{2}{\begin{bmatrix}0\\0\\0\\1\end{bmatrix}}$$={\begin{bmatrix}1&0\\-3&0\\0&0\\0&1\end{bmatrix}}{\begin{bmatrix}s_{1}\\s_{2}\end{bmatrix}}$

Using the equation from part (c), we have

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

${\begin{bmatrix}8&4&2&1\\1&1&1&1\\0&0&0&1\end{bmatrix}}{\begin{bmatrix}1&0\\-3&0\\0&0\\0&1\end{bmatrix}}{\vec {s}}={\begin{bmatrix}2\\2\\1\end{bmatrix}}$

Therefore,

$C={\begin{bmatrix}-4&1\\-2&1\\0&1\end{bmatrix}}$ and ${\vec {c}}={\vec {b}}={\begin{bmatrix}2\\2\\1\end{bmatrix}}$.

Notice that for there to be a solution we need ${\vec {c}}$ 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 $N(C^{T})$ ${\begin{bmatrix}-4&-2&0\\1&1&1\end{bmatrix}}{\vec {x}}=0\rightarrow {\begin{bmatrix}2&1&0\\0&1&2\end{bmatrix}}{\vec {x}}=0$

${\vec {x}}={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}1\\-2\\1\end{bmatrix}}x_{3}$

Since ${\begin{bmatrix}1\\-2\\1\end{bmatrix}}\cdot {\begin{bmatrix}2\\2\\1\end{bmatrix}}=2-4+1\neq 0$

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