The linear transformation \(\mathrm{T} : \mathbb{R}^4 \to \mathbb{R}^4\) is represented by the matrix \(\mathbf{M}\), where $$\mathbf{M} = \begin{pmatrix} 1 & -2 & -3 & 1 \\ 3 & -5 & -7 & 7 \\ 5 & -9 & -13 & 9 \\ 7 & -13 & -19 & 11 \end{pmatrix}.$$ Find the rank of \(\mathbf{M}\) and a basis for the null space of \(\mathrm{T}\). [6] The vector \(\begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}\) is denoted by \(\mathbf{e}\). Show that there is a solution of the equation \(\mathbf{M}\mathbf{x} = \mathbf{M}\mathbf{e}\) of the form $$\mathbf{x} = \begin{pmatrix} a \\ b \\ -1 \\ -1 \end{pmatrix}, \text{ where the constants } a \text{ and } b \text{ are to be found.}$$ [4]

Question 7:
7 | 1 −2 −3 1 1 −2 −3 1 1 −2 −3 1
     
3 −5 −7 7 0 1 2 4 0 1 2 4
~ ~
     
5 −9 −13 9 0 1 2 4 0 0 0 0
     
     
7 −13 −19 11 0 1 2 4 0 0 0 0
r(M) = 4 – 2 = 2
x−2y−3z+t =0
y+2z+4t =0
E.g. Set z = λ and t = µ ⇒ y=−2λ−4µ and x=−λ−9µ
−1 −9
   
−2 −4
⇒Basis is   ,  
1 0
   
    0     1    
1 −1 −9  a 
       
2 −2 −4  b 
x= +λ +µ =
       
3 1 0 −1
       
       
4  0   1  −1
Solving: λ =−4 and µ =−5
⇒a=50, b=30. | M1A1
A 1
M1
M1
A1
[6]
M1
M1 A1
A1
[4]
Total
10

The linear transformation $\mathrm{T} : \mathbb{R}^4 \to \mathbb{R}^4$ is represented by the matrix $\mathbf{M}$, where
$$\mathbf{M} = \begin{pmatrix} 1 & -2 & -3 & 1 \\ 3 & -5 & -7 & 7 \\ 5 & -9 & -13 & 9 \\ 7 & -13 & -19 & 11 \end{pmatrix}.$$

Find the rank of $\mathbf{M}$ and a basis for the null space of $\mathrm{T}$. [6]

The vector $\begin{pmatrix} 1 \\ 2 \\ 3 \\ 4 \end{pmatrix}$ is denoted by $\mathbf{e}$. Show that there is a solution of the equation $\mathbf{M}\mathbf{x} = \mathbf{M}\mathbf{e}$ of the form
$$\mathbf{x} = \begin{pmatrix} a \\ b \\ -1 \\ -1 \end{pmatrix}, \text{ where the constants } a \text{ and } b \text{ are to be found.}$$ [4]

\hfill \mbox{\textit{CAIE FP1 2015 Q7 [10]}}