# How do you solve ((-3, -2), (1, 1))X=((-8, -1),(6,0))?

May 30, 2016

$\textcolor{g r e e n}{\implies X = \left(\begin{matrix}- 4 & + 1 \\ + 10 & - 1\end{matrix}\right)}$

#### Explanation:

Deliberately not using the shortcut method of using the determinant.

Write as $A X = B$

Then ${A}^{- 1} A X = {A}^{- 1} B$

But ${A}^{- 1} A = I$

$\implies X = {A}^{- 1} B$

$\textcolor{red}{\text{Note that the order of the matrices is important}}$
$\textcolor{m a \ge n t a}{{A}^{- 1} A \ne A {A}^{- 1}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine } {A}^{- 1}}$

$\left(\begin{matrix}- 3 & - 2 \text{ |" 1 & 0 \\ " "1 & 1 color(white)(..)" |} 0 & 1\end{matrix}\right)$
'...............................................

$R o w \left(1\right) + 3 R o w \left(2\right)$

$\left(\begin{matrix}0 & 1 \text{ |" 1 & 3 \\ 1 & 1 " |} 0 & 1\end{matrix}\right)$
'....................................................

$R o w \left(2\right) - R o w \left(1\right)$

$\left(\begin{matrix}0 & \textcolor{w h i t e}{\ldots} 1 \text{ |"color(white)(..) 1 & 3 \\ 1 & color(white)(..) 0 " |} - 1 & - 2\end{matrix}\right)$
'................................................

Reverse the order of the rows

$\left(\begin{matrix}1 & \textcolor{w h i t e}{. .} 0 \text{ |"-1 & -2 \\ 0 & color(white)(.)1" |} \textcolor{w h i t e}{. .} 1 & 3\end{matrix}\right)$
'.......................................................

$\textcolor{b r o w n}{\implies {A}^{- 1} = \left(\begin{matrix}- 1 & - 2 \\ 1 & 3\end{matrix}\right)}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\implies X = {A}^{- 1} B$

$\implies X = \left(\begin{matrix}- 1 & - 2 \\ 1 & 3\end{matrix}\right) \left(\begin{matrix}- 8 & - 1 \\ 6 & 0\end{matrix}\right)$

$\text{Let } X = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

$a = \left[\left(- 1\right) \times \left(- 8\right)\right] \text{ "+" } \left[\left(- 2\right) \times \left(6\right)\right] = - 4$

$b = \left[\left(- 1\right) \times \left(- 1\right)\right] \text{ "+" } \left[\left(- 2\right) \times \left(0\right)\right] = + 1$

$c = \left[\left(1\right) \times \left(- 8\right)\right] \text{ "+" } \left[\left(3\right) \times \left(6\right)\right] = + 10$

$d = \left[\left(1\right) \times \left(- 1\right)\right] \text{ "+" } \left[\left(3\right) \times \left(0\right)\right] = - 1$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{g r e e n}{\implies X = \left(\begin{matrix}- 4 & + 1 \\ + 10 & - 1\end{matrix}\right)}$