(i)     Find the row rank of \(M\).

(ii)     Hence or otherwise find the kernel of \(T\).

(i)     State the column rank of \(M\).

(ii)     Find the basis for the range of this transformation.

(i)     row reduction gives \(\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&7&7 \\ 0&3&3 \end{array}} \right) \to \left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)\left( { \to \left( {\begin{array}{*{20}{c}} 1&0&{ - 1} \\ 0&1&1 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)} \right)\)     (M1)A1

hence row rank is \(2\)     A1

Note: Accept the argument that Column 2 = Column 1 + Column 3

(ii)     to find the kernel \(\left( {\begin{array}{*{20}{c}} 1&2&1 \\ 0&3&3 \\ 0&0&0 \\ 0&0&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \\ 0 \\ 0 \end{array}} \right)\)     M1

Note: Allow the use of the original matrix

\(x + 2y + z = 0\)

\(3y + 3z = 0\)     A1

let \(z = \lambda \)     M1

hence \(y =  - \lambda ,{\text{ }}x = \lambda \)

the kernel is therefore \(\left[ {\begin{array}{*{20}{c}} \lambda  \\ { - \lambda } \\ \lambda \end{array}} \right]\)     A1

(i)     column rank is \(2\)     A1

(ii)     a basis for the range is defined by two independent vectors     (M1)

therefore a basis for the range is for example, \(\left[ {\begin{array}{*{20}{c}} 1 \\ 2 \\ { - 3} \\ 1 \end{array}} \right]\) and \(\left[ {\begin{array}{*{20}{c}} 2 \\ 7 \\ 1 \\ 5 \end{array}} \right]\)     A2

Many solutions to this question suggested that the topic had not been adequately covered in many centres so that solutions were either good or virtually non existent. Most successful candidates used their calculator to perform the row reduction.

Many solutions to this question suggested that the topic had not been adequately covered in many centres so that solutions were either good or virtually non existent. Most successful candidates used their calculator to perform the row reduction.