The images of points (2, -3) and (4, 5) under a... | WAEC SSCE - Further Mathematics - 2011 (Essay)

Question 1 Report

The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the :

(a) matrix A ; (b) inverse of A ; (c) point whose image is (-1, 1).

Answer

Let \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The two mappings give:

\(A\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\Rightarrow 2a-3b=3,\ 2c-3d=4\)

\(A\begin{pmatrix}4\\5\end{pmatrix}=\begin{pmatrix}5\\6\end{pmatrix}\Rightarrow 4a+5b=5,\ 4c+5d=6\)

(a) Solve for \(a,b\): from \(2a-3b=3\) and \(4a+5b=5\), eliminating \(a\) gives \(11b=-1\Rightarrow b=-\tfrac{1}{11}\), then \(a=\tfrac{15}{11}\). Solve for \(c,d\): \(11d=-2\Rightarrow d=-\tfrac{2}{11}\), then \(c=\tfrac{19}{11}\).

\[A=\frac{1}{11}\begin{pmatrix}15&-1\\19&-2\end{pmatrix}\]

(b) \(\det A=\dfrac{(15)(-2)-(-1)(19)}{11^2}=\dfrac{-11}{121}=-\dfrac{1}{11}\).

\[A^{-1}=\frac{1}{\det A}\cdot\frac{1}{11}\begin{pmatrix}-2&1\\-19&15\end{pmatrix}=\begin{pmatrix}2&-1\\19&-15\end{pmatrix}\]

(c) The point mapping to \((-1,1)\) is \(A^{-1}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}2(-1)-1(1)\\19(-1)-15(1)\end{pmatrix}=\begin{pmatrix}-3\\-34\end{pmatrix}\).

So the point is \((-3,-34)\).

Read lesson note on Matrices And Linear Transformation (WAEC)

Answer Details

Let \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\). The two mappings give:

\(A\begin{pmatrix}2\\-3\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\Rightarrow 2a-3b=3,\ 2c-3d=4\)

\(A\begin{pmatrix}4\\5\end{pmatrix}=\begin{pmatrix}5\\6\end{pmatrix}\Rightarrow 4a+5b=5,\ 4c+5d=6\)

\[A=\frac{1}{11}\begin{pmatrix}15&-1\\19&-2\end{pmatrix}\]

(b) \(\det A=\dfrac{(15)(-2)-(-1)(19)}{11^2}=\dfrac{-11}{121}=-\dfrac{1}{11}\).

\[A^{-1}=\frac{1}{\det A}\cdot\frac{1}{11}\begin{pmatrix}-2&1\\-19&15\end{pmatrix}=\begin{pmatrix}2&-1\\19&-15\end{pmatrix}\]

(c) The point mapping to \((-1,1)\) is \(A^{-1}\begin{pmatrix}-1\\1\end{pmatrix}=\begin{pmatrix}2(-1)-1(1)\\19(-1)-15(1)\end{pmatrix}=\begin{pmatrix}-3\\-34\end{pmatrix}\).

So the point is \((-3,-34)\).

Read lesson note on Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

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The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the : (a) matrix A ; (b) inverse of A ; (c)...

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The images of points (2, -3) and (4, 5) under a linear transformation A are (3, 4) and (5, 6) respectively. Find the : (a) matrix A ; (b) inverse of A ; (c)...

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