A-LevelFurther MathematicsMatrices9231/12 · May/June 2011 · Q36 Marks
The linear transformation \( T : \mathbb{R}^4 \to \mathbb{R}^4 \) is represented by the matrix \( M = \begin{pmatrix} 1 & 3 & -2 & 4 \\ 5 & 15 & -9 & 19 \\ -2 & -6 & 3 & -7 \\ 3 & 9 & -5 & 11 \end{pmatrix} \).
- (i)Find the rank of M.[3]
- (ii)Obtain a basis for the null space of T.[3]
Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- [M1A1] Reduces matrix to echelon form. \( \begin{pmatrix} 1 & 3 & -2 & 4 \\ 5 & 15 & -9 & 19 \\ -2 & -6 & 3 & -7 \\ 3 & 9 & -5 & 11 \end{pmatrix} \to \begin{pmatrix} 1 & 3 & -2 & 4 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \)
- [A1] Uses rank = dimension – nullity \( r(A) = 4 - 2 = 2 \)
- [B1] Obtains a set of equations. \( x+3y-2z+ 4t = 0 \) \( z-t=0 \)
- [M1A1] Finds basis vectors. Basis is \( \begin{pmatrix} -2 \\ 0 \\ 1 \\ 1 \end{pmatrix} \), \( \begin{pmatrix} -3 \\ 1 \\ 0 \\ 0 \end{pmatrix} \) (OE)
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