Challenging +1.2 This question requires computing two determinants with parameter k, finding the area of the original triangle, applying the area scale factor property (|det(M)|), then solving an inequality involving absolute values. While it combines multiple steps (determinant calculation, area formula, absolute value inequality), each component is standard Further Maths material with no novel insight required. The algebraic manipulation is moderate but straightforward.
16.
$$\begin{gathered}
M _ { 1 } = \left( \begin{array} { c c }
2 k - 9 & 5 - k \\
- k & k - 2
\end{array} \right) \\
M _ { 2 } = \left( \begin{array} { c c }
5 & 1 \\
2 k - 3 & k - 3
\end{array} \right) \\
k \in \mathbb { R }
\end{gathered}$$
Matrices \(M _ { 1 }\) and \(M _ { 2 }\) represent transformations \(T _ { 1 }\) and \(T _ { 2 }\) respectively.
\(\Delta\) is a triangle in the \(x y\)-plane with vertices at \(( 0,0 ) , ( 4,0 )\) and \(( 3,2 )\).
The image of \(\Delta\) under \(T _ { 1 }\) is \(\Delta _ { 1 }\) and the image of \(\Delta\) under \(T _ { 2 }\) is \(\Delta _ { 2 }\).
The area of \(\Delta _ { 2 }\) is greater than the area of \(\Delta _ { 1 }\).
Find the range of possible values of \(k\). [0pt]
[0pt]
[0pt]
[0pt]
[0pt]
[0pt]
16.
$$\begin{gathered}
M _ { 1 } = \left( \begin{array} { c c }
2 k - 9 & 5 - k \\
- k & k - 2
\end{array} \right) \\
M _ { 2 } = \left( \begin{array} { c c }
5 & 1 \\
2 k - 3 & k - 3
\end{array} \right) \\
k \in \mathbb { R }
\end{gathered}$$
Matrices $M _ { 1 }$ and $M _ { 2 }$ represent transformations $T _ { 1 }$ and $T _ { 2 }$ respectively.\\
$\Delta$ is a triangle in the $x y$-plane with vertices at $( 0,0 ) , ( 4,0 )$ and $( 3,2 )$.\\
The image of $\Delta$ under $T _ { 1 }$ is $\Delta _ { 1 }$ and the image of $\Delta$ under $T _ { 2 }$ is $\Delta _ { 2 }$.\\
The area of $\Delta _ { 2 }$ is greater than the area of $\Delta _ { 1 }$.\\
Find the range of possible values of $k$.\\[0pt]
\\[0pt]
\\[0pt]
\\[0pt]
\\[0pt]
\\[0pt]
\hfill \mbox{\textit{SPS SPS FM Pure 2024 Q16 [9]}}