0606 P12 - Mar 2018 - Q7 - 9 marks
8392
(a) Find the values of \(a\) for which the determinant of \(\begin{pmatrix}2a&1\\4a&a\end{pmatrix}\) is \(6-3a\).
(b) Let \(A=\begin{pmatrix}2&1\\3&4\end{pmatrix}\) and \(B=\begin{pmatrix}2&0\\-3&5\end{pmatrix}\).
(i) Find \(A^{-1}\).
(ii) Hence find the matrix \(C\) such that \(AC=B\).
(c) Find the \(2\times2\) matrix \(D\) such that \(4D+3I=O\), where \(I\) is the identity matrix and \(O\) is the zero matrix.
Solution
Answer: \(a=2\) or \(a=-\dfrac32\); \(A^{-1}=\dfrac15\begin{pmatrix}4&-1\\-3&2\end{pmatrix}\); \(C=\dfrac15\begin{pmatrix}11&-5\\-12&10\end{pmatrix}\); \(D=\begin{pmatrix}-\dfrac34&0\\0&-\dfrac34\end{pmatrix}\).
The determinant of \(\begin{pmatrix}2a&1\\4a&a\end{pmatrix}\) is
\((2a)(a)-1(4a)=2a^2-4a.\)
This is equal to \(6-3a\), so
\(2a^2-4a=6-3a.\)
Hence
\(2a^2-a-6=0.\)
Factorising,
\((2a+3)(a-2)=0,\)
so
\(a=-\dfrac32\quad\text{or}\quad a=2.\)
For \(A=\begin{pmatrix}2&1\\3&4\end{pmatrix}\),
\(\det A=2\cdot4-1\cdot3=5.\)
Therefore
\(A^{-1}=\dfrac15\begin{pmatrix}4&-1\\-3&2\end{pmatrix}.\)
If \(AC=B\), then
\(C=A^{-1}B.\)
So
\(C=\dfrac15\begin{pmatrix}4&-1\\-3&2\end{pmatrix}\begin{pmatrix}2&0\\-3&5\end{pmatrix}=\dfrac15\begin{pmatrix}11&-5\\-12&10\end{pmatrix}.\)
For \(4D+3I=O\),
\(4D=-3I,\)
so
\(D=-\dfrac34I=\boxed{\begin{pmatrix}-\dfrac34&0\\0&-\dfrac34\end{pmatrix}}.\)
