Exam-Style Problem

9231 P11 - Nov 2020 - Q1 - 9 marks

5801

1 The matrix \(\mathbf{M}\) is given by \(\mathbf{M}=\left(\begin{array}{ll}1 & b \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}a & 0 \\ 0 & 1\end{array}\right)\), where \(a\) and \(b\) are positive constants.
(a) The matrix \(\mathbf{M}\) represents a sequence of two geometrical transformations.

State the type of each transformation, and make clear the order in which they are applied.

The unit square in the \(x-y\) plane is transformed by \(\mathbf{M}\) onto parallelogram \(O P Q R\).
(b) Find, in terms of \(a\) and \(b\), the matrix which transforms parallelogram \(O P Q R\) onto the unit square.

It is given that the area of \(O P Q R\) is \(2 \mathrm{~cm}^{2}\) and that the line \(x+3 y=0\) is invariant under the transformation represented by \(\mathbf{M}\).
(c) Find the values of \(a\) and \(b\).

Solution Checked by expert

Answer:

(a) A one-way stretch in the \(x\)-direction with scale factor \(a\), followed by a shear parallel to the \(x\)-axis with shear factor \(b\).
(b) The required matrix is \(\mathbf M^{-1}=\dfrac{1}{a}\begin{pmatrix}1&-b\\0&a\end{pmatrix}\).
(c) \(a=2\) and \(b=3\).

(a) The matrix is written as \(\mathbf M=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}\). For column vectors, the right-hand matrix acts first.
The matrix \(\begin{pmatrix}a&0\\0&1\end{pmatrix}\) maps \((x,y)\) to \((ax,y)\), so it is a one-way stretch in the \(x\)-direction with scale factor \(a\).
The matrix \(\begin{pmatrix}1&b\\0&1\end{pmatrix}\) maps \((x,y)\) to \((x+by,y)\), so it is a shear parallel to the \(x\)-axis with shear factor \(b\).
Therefore the stretch is applied first, followed by the shear.
(b) First multiply the matrices:
\(\mathbf M=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\).
The matrix which maps the transformed parallelogram back to the unit square is the inverse matrix \(\mathbf M^{-1}\).
For \(\begin{pmatrix}a&b\\0&1\end{pmatrix}\), the determinant is \(a\), so
\(\mathbf M^{-1}=\dfrac{1}{a}\begin{pmatrix}1&-b\\0&a\end{pmatrix}\).
(c) The area scale factor of a matrix transformation is the absolute value of its determinant. Since \(a\) and \(b\) are positive,
\(\det\mathbf M=a\).
The unit square has area \(1\), and its image has area \(2\), so \(a=2\).
Now \(\mathbf M\) maps \((x,y)\) to \((X,Y)\), where
\(\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\y\end{pmatrix}\).
The line \(x+3y=0\) has \(y=-\dfrac{x}{3}\). Substituting this into the image coordinates gives
\(X=ax+b\left(-\dfrac{x}{3}\right)=\left(a-\dfrac{b}{3}\right)x\), and \(Y=-\dfrac{x}{3}\).
For the line to be invariant, the image point must also lie on the same line, so \(X+3Y=0\):
\(\left(a-\dfrac{b}{3}\right)x+3\left(-\dfrac{x}{3}\right)=0\).
Thus \(\left(a-\dfrac{b}{3}-1\right)x=0\), so \(a-\dfrac{b}{3}=1\).
Using \(a=2\),
\(2-\dfrac{b}{3}=1\), so \(b=3\).
Therefore \(a=2\) and \(b=3\).