# Matrices

• Jan 25th 2013, 08:17 AM
MMCS
Matrices
A transformation in the plane is represented by a 2 x 2 matrix.
Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2)

Can someone help me to find the transformation matrix for this?
I dont have any working to show as i dont know where to start

Thanks
• Jan 25th 2013, 09:01 AM
Soroban
Re: Matrices
Hello, MMCS!

Quote:

A transformation in the plane is represented by a 2 x 2 matrix.
Under this transformation, the point (7,7) is transformed to (-5,-6) and the point (-2,8) is transformed to (-4,2).
Find the transformation matrix.

You can start by letting the transformation matrix be: . $A \;=\;\begin{bmatrix} a&b \\ c&d\end{bmatrix}$

Then: . $(7,7)\begin{bmatrix}a&b\\c&d\end{bmatrix} \;=\;(\text{-}5,\text{-}6) \quad\Rightarrow\quad \begin{Bmatrix}7a + 7c &=& \text{-}5 \\ 7b + 7d &=& \text{-}6 \end{Bmatrix}$

Also: . $(\text{-}2,8)\begin{bmatrix}a&b\\c&d\end{bmatrix}\;=\;( \text{-}4,2) \quad\Rightarrow\quad \begin{Bmatrix}\text{-}2a + 8c &=& \text{-}4 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}$

We have two systems of equations:

. . . . . . . . . . . $\begin{Bmatrix}7a + 7c &=& \text{-}5 \\ \text{-}2a+8c &=& \text{-}4\end{Bmatrix} \qquad\begin{Bmatrix}7b+7d &=& \text{-}6 \\ \text{-}2b + 8d &=& 2 \end{Bmatrix}$

Their solutions are: . $\begin{Bmatrix} a&=& \text{-}\frac{6}{35} \\ \\[-4mm] c &=& \text{-}\frac{19}{35} \end{Bmatrix} \qquad \begin{Bmatrix}b &=& \text{-}\frac{31}{35} \\ \\[-4mm] d &=& \frac{11}{35} \end{Bmatrix}$

Therefore: . $A \;=\;\begin{bmatrix}\text{-}\frac{6}{35} & \text{-}\frac{31}{35} \\ \\[-3mm] \text{-}\frac{19}{35} & \frac{1}{35} \end{bmatrix}$
• Jan 25th 2013, 09:30 AM
ILikeSerena
Re: Matrices
Here's an alternative method:

We have:

$A \begin{pmatrix}7 \\ 7\end{pmatrix} = \begin{pmatrix}-5 \\ -6\end{pmatrix} \qquad A \begin{pmatrix}-2 \\ 8\end{pmatrix} = \begin{pmatrix}-4 \\ 2\end{pmatrix}$

Therefore:

$A \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix} = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix}$

$A = \begin{pmatrix}-5 & -4 \\ -6 & 2\end{pmatrix} \begin{pmatrix}7 & -2 \\ 7 & 8\end{pmatrix}^{-1}$
• Jan 25th 2013, 09:45 AM
HallsofIvy
Re: Matrices
Yet a third way: (1, 0)= a(7, 7)+ b(-2, 8) for some a and b. That gives the two equations 7a- 2b= 1 and 7a+ 8b= 0. If we subtract the first equation from the second we get 10b= -1 so that b= -1/10. Putting that back into the first equation, 7a+ 2/10= 1 so 7a= 8/10 and a= 8/70. That is (1, 0)= (8/70)(7, 7)+ (-1/10)(-2, 8) and then A(1, 0)= (8/70)A(7, 7)+ (-1/10)A(-2, 8)= (8/70)(-5, -6)+ (-1/10)(-4, 2)= (8/70, -31/35). The point of reducing to (1, 0) is that $\begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}$. That is the first column of the matrix is whatever A maps (1, 0) to. This tells us that the first column of the matrix is $\begin{bmatrix}\frac{8}{70} \\ -\frac{31}{35}\end{bmatrix}$.

Now, find a, b such that a(7, 7)+ b(-2, 8)= (0, 1) to find the second column.