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http://studyguide.pk/Past%20Papers/CIE/International%20A%20And%20AS%20Level/9231%20-%20Further%20Mathematics/9231_s03_qp_1.pdf

I would like to know the method to answer question 8. I have been having difficulties with finding the basis for the null space of a matrix although I know the basic method. Also, in the second part of the question I have no clue as to how I should proceed to find p and q. I would appreciate it if someone could help me. Thanks.

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Row reduction leads to the form: $$\begin{bmatrix} 1&-1&-2&3\\ 0&1&3&5\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}$$ hence the matrix has rank $2$.

To find a basis for the null space, proceed with row reduction upwards. One gets: $$\begin{bmatrix} 1&0&1&8\\ 0&1&3&5\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}$$ To solve for $A\,\mathbf x=0$, this last form shows that the main unknowns are $x,y\mkern1.5mu$; they can be calculated in function of $z$ and $t$. Indeed: $$\begin{cases}x=-z-8t,\\y=-3z-5t\end{cases}, \text{whence}\enspace \begin{bmatrix} x\\y\\z\\t \end{bmatrix}=\begin{bmatrix} -z-3t\\-3z-3t\\z\\t \end{bmatrix}=-z\begin{bmatrix} 1\\3\\-1\\0 \end{bmatrix} -t\begin{bmatrix} 3\\3\\0\\-1 \end{bmatrix}.$$

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