# Finding a two-by-two matrix such that…

I need to find a matrix which would solve for the following: $$\begin{bmatrix}4 & 7\\4 & 2 \end{bmatrix} * \begin{bmatrix}- & -\\- &- \end{bmatrix} = \begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix}$$

I tried using the inverse which gave me $\begin{bmatrix}\frac{2}{-20} & \frac{-7}{-20}\\\frac{-4}{-20} & \frac{4}{-20} \end{bmatrix}$

Also attempted making $\Bigg [\begin{matrix}4 & 7\\4 & 2 \end{matrix}\Bigg | \begin{matrix}1 & 0\\0 & 1 \end{matrix}\Bigg ]$

which gave me the following: $\Bigg [\begin{matrix}1 & 0\\0 & 1 \end{matrix}\Bigg | \begin{matrix}\frac17 & 0\\\frac1{20} & \frac1{-5} \end{matrix}\Bigg ]$

I am guessing I did something wrong at some step but Cannot seem to figure out where!

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First thing to do is to check to see which (if either) answer is right. That's easy --- just do the multiplication, and see whether you get the identity matrix. –  Gerry Myerson Feb 4 '13 at 11:53
WolframAlpha says the inverse works: wolframalpha.com/input/?i=[[4%2C7]%2C[4%2C2]]*[[-2%2F20%2C7%2F20]%2C‌​[4%2F20%2C-4%2F20]] –  Inkbug Feb 4 '13 at 11:55
Thank you, I must of typed it wrong in the system ... wasted so much time on this. –  Reza M. Feb 4 '13 at 12:01

I have a trick for you. Let $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ be any matrix. Then its inverse is $$\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a\end{bmatrix}$$
Where "any matrix" means "any invertible $2\times2$ matrix." –  Gerry Myerson Feb 4 '13 at 12:21