# Thread: matrix

1. ## matrix

Solve the matrix X:

2. Originally Posted by Mr_Green
Solve the matrix X:

$\displaystyle X \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right ) - \left ( \begin{array}{cc} -5 & 2 \\ 6 & 3 \end{array} \right ) = \left ( \begin{array}{cc} -2 & -9 \\ -5 & -22 \end{array} \right )$

$\displaystyle X \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right ) - \left ( \begin{array}{cc} -5 & 2 \\ 6 & 3 \end{array} \right ) + \left ( \begin{array}{cc} -5 & 2 \\ 6 & 3 \end{array} \right )= \left ( \begin{array}{cc} -2 & -9 \\ -5 & -22 \end{array} \right ) + \left ( \begin{array}{cc} -5 & 2 \\ 6 & 3 \end{array} \right )$

$\displaystyle X \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right ) = \left ( \begin{array}{cc} -7 & -7 \\ 1 & -19 \end{array} \right )$

$\displaystyle X \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right ) \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right )^{-1} = \left ( \begin{array}{cc} -7 & -7 \\ 1 & -19 \end{array} \right ) \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right )^{-1}$

Now,
$\displaystyle \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right )^{-1} = \frac{1}{ad-bc} \left ( \begin{array}{cc} d & -b \\ -c & a \end{array} \right )$
so:

$\displaystyle \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right )^{-1} = \frac{1}{3 \cdot 4 - -1 \cdot 2} \left ( \begin{array}{cc} 4 & 1 \\ -2 & 3 \end{array} \right )$ $\displaystyle = \frac{1}{14} \left ( \begin{array}{cc} 4 & 1 \\ -2 & 3 \end{array} \right )$

Thus
$\displaystyle X \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right ) \cdot \left ( \begin{array}{cc} 3 & -1 \\ 2 & 4 \end{array} \right )^{-1} = \left ( \begin{array}{cc} -7 & -7 \\ 1 & -19 \end{array} \right ) \cdot \frac{1}{14} \left ( \begin{array}{cc} 4 & 1 \\ -2 & 3 \end{array} \right )$

$\displaystyle X = \left ( \begin{array}{cc} -1 & -2 \\ 3 & -4 \end{array} \right )$

-Dan