Check what the transformation does for each vector in the standard basis of R²<em> </em>:
<em>T</em> (1, 0) = (5, 2)
<em>T</em> (0, 1) = (-4, 0)
Now compute the weights <em>a</em>, <em>b</em> and <em>c</em>, <em>d</em> such that
<em>a</em> <em>T</em> (1, 0) + <em>b</em> <em>T</em> (0, 1) = (-2, 1)
<em>c</em> <em>T</em> (1, 0) + <em>d</em> <em>T</em> (0, 1) = (-1, 1)
![\left[\begin{array}{cc|c}5&-4&-2\\2&0&1\end{array}\right]\sim\left[\begin{array}{cc|c}1&0&\frac12\\\\0&1&\frac98\end{array}\right]\implies a=\dfrac12,b=\dfrac98](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%26-4%26-2%5C%5C2%260%261%5Cend%7Barray%7D%5Cright%5D%5Csim%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%260%26%5Cfrac12%5C%5C%5C%5C0%261%26%5Cfrac98%5Cend%7Barray%7D%5Cright%5D%5Cimplies%20a%3D%5Cdfrac12%2Cb%3D%5Cdfrac98)
![\left[\begin{array}{cc|c}5&-4&-1\\2&0&1\end{array}\right]\sim\left[\begin{array}{cc|c}1&0&\frac12\\\\0&1&\frac78\end{array}\right]\implies c=\dfrac12,d=\dfrac78](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%26-4%26-1%5C%5C2%260%261%5Cend%7Barray%7D%5Cright%5D%5Csim%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%260%26%5Cfrac12%5C%5C%5C%5C0%261%26%5Cfrac78%5Cend%7Barray%7D%5Cright%5D%5Cimplies%20c%3D%5Cdfrac12%2Cd%3D%5Cdfrac78)
Then the matrix <em>A'</em> is
![A'=\begin{bmatrix}\frac12&\frac12\\\\\frac98&\frac78\end{bmatrix}](https://tex.z-dn.net/?f=A%27%3D%5Cbegin%7Bbmatrix%7D%5Cfrac12%26%5Cfrac12%5C%5C%5C%5C%5Cfrac98%26%5Cfrac78%5Cend%7Bbmatrix%7D)