FP1 2017 5 Edexcel

(ia)

We have:

AB = \begin{pmatrix}p & 2 \\ 3 & p\end{pmatrix} \begin{pmatrix}-5 & 4 \\ 6 & -5\end{pmatrix} = \begin{pmatrix}-5p + 12 & 4p - 10 \\ -15 + 6p & -5p + 12\end{pmatrix}

(ib)

We have:

\begin{align*}AB + 2A & = \begin{pmatrix}-5p + 12 & 4p - 10 \\ -15 + 6p & -5p + 12\end{pmatrix} + \begin{pmatrix}2p & 4 \\ 6 & 2p\end{pmatrix} \\ & = \begin{pmatrix}-3p + 12 & 4p - 6 \\ 6p - 9 & -3p + 12\end{pmatrix}\end{align*}

We need to have 4p - 6 = 6p - 9 = 0, so p = \dfrac 6 4 = \dfrac 3 2.

Then:

\begin{align*}\begin{pmatrix}-3p + 12 & 4p - 6 \\ 6p - 9 & -3p + 12\end{pmatrix} &= \begin{pmatrix}-3 \left(\frac 3 2\right) + 12 & 0 \\ 0 & -3\left(\frac 3 2\right) + 12\end{pmatrix} \\ & = \begin{pmatrix}\frac {15} 2 & 0 \\ 0 & \frac {15} 2\end{pmatrix} \\ & = \frac {15} 2 I\end{align*}

So k = \dfrac {15} 2.

(ii)

Note that \det M gives the area scale factor when M is applied to some plane figure. (ie. the area will increase by a factor of |\det M|) Note that we have \det M = 2a + 9, and:

\displaystyle 15|2a +9| = 270

So:

\displaystyle |2a + 9| = 18

So either:

\displaystyle 2a + 9 =18

giving a = \dfrac 9 2 or:

\displaystyle 2a + 9 = -18

giving a = -\dfrac {27} 2.

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