Solve the following equations using the Matrix Inversion method.
$2x - 3y + 6 = 0$ and $6x + y + 8 = 0$.

Hint: If two equations are in the form of $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$, then matrix inversion method gives us ${A^{ - 1}}$ to $AX = B$ where $A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right]$ , $B=\left[ \begin{matrix}
   {{c}_{1}} \\
   {{c}_{2}} \\
\end{matrix} \right]$
and $X = \left[ \begin{matrix}
   {{x}_{1}} \\
   {{x}_{2}} \\
\end{matrix} \right]$ . We first convert the equations in standard form $ax + by = c$and then find the determinant and adjoint of $A$. Using the Matrix Inversion formula, we find the value of $X$ in terms of a matrix.

Formula used:
The given equations are $2x - 3y + 6 = 0$ and $6x + y + 8 = 0$
To use the matrix inversion method, we have to first convert the equations in standard form as $ax + by = c$.
The equations in standard form are $2x - 3y = - 6$ and $6x + y = - 8$
The matrix equation is $AX = B$
Where $A = \left[ {\begin{array}{*{20}{c}}
  {{a_1}}&{{b_1}} \\
  {{a_2}}&{{b_2}}
\end{array}} \right]$ , $B = \left[ {\begin{array}{*{20}{c}}
  {{c_1}}\\
  {{c_2}}
\end{array}} \right]$ and $X = \left[ {\begin{array}{*{20}{c}}
  {{x_1}}\\
  {{x_2}} \end{array}} \right]$
The values of equation in matrix form are
$A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  6&1
\end{array}} \right]$ , $B = \left[ {\begin{array}{*{20}{c}}
  -6\\
  -8
\end{array}} \right]$ and $X = \left[ {\begin{array}{*{20}{c}}
  x\\
  y \end{array}} \right]$
Using the Matrix Inversion method,
$\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  6&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x\\
  y \end{array}} \right]= \left[ {\begin{array}{*{20}{c}}
  -6\\
  -8
\end{array}} \right]$
The matrix equation is $AX = B$
Pre multiplying the above matrix by ${A^{ - 1}}$ on both sides of equation we get,
${A^{ - 1}}AX = {A^{ - 1}}B$
We know, ${A^{ - 1}}A = I$ and $IX = X$,
Therefore, $IX = {A^{ - 1}}B$
$X = {A^{ - 1}}B$
Now we will find the determinant of matrix $A$.
$A = \left| {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  6&1
\end{array}} \right|$
$\left| A \right| = (2 \times 1) - ( - 3 \times 6)$
$\left| A \right| = 2 + (3 \times 6)$
$\left| A \right| = 2 + 18$
$\left| A \right| = 20$
If $A = \left[ {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right]$ then adjoint of $A = \left[ {\begin{array}{*{20}{c}}
  d&{ - b} \\
  { - c}&a
\end{array}} \right]$
Here $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  6&1
\end{array}} \right]$ then adjoint of $A = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  { - 6}&2
\end{array}} \right]$
We know, ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj(A)$
Substituting the values in the formula,
${A^{ - 1}} = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  1&3 \\
  { - 6}&2
\end{array}} \right]$
Now we have ${A^{ - 1}} = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  1&3 \\
  { - 6}&2
\end{array}} \right]$, $B = \left[ {\begin{array}{*{20}{c}}
  -6 \\
  -8
\end{array}} \right]$ and $X = \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]$
So, substituting these values in the equation $X = {A^{ - 1}}B$,
$X = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  1&3 \\
  { - 6}&2
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  -6 \\
  -8
\end{array}} \right]$
Multiplying the matrices,
$X = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  (1 \times - 6) + (3 \times - 8) \\
  ( - 6 \times - 6) + (2 \times - 8)
\end{array}} \right]$
Solving the arithmetic,
$X = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  -6-24 \\
  36-16
\end{array}} \right]$
Solving the arithmetic,
$X = \dfrac{1}{{20}}\left[ {\begin{array}{*{20}{c}}
  -30 \\
  20
\end{array}} \right]$
Taking the determinant inside the matrix,
$X = \left[ {\begin{array}{*{20}{c}}
  \dfrac{-30}{{20}} \\
  \dfrac{20}{{20}}
\end{array}} \right]$
Simplifying the elements,
$X = \left[ {\begin{array}{*{20}{c}}
  \dfrac{-3}{{2}} \\
  1
\end{array}} \right]$
$X = \left[ {\begin{array}{*{20}{c}}
  \dfrac{-3}{{2}} \\
  1
\end{array}} \right]= \left[ {\begin{array}{*{20}{c}}
  x \\
  y
\end{array}} \right]$
Therefore, we get, $x = \dfrac{{ - 3}}{2}$ and $y = 1$ .
The solution of the matrices is $x = \dfrac{{ - 3}}{2}$ and $y = 1$.

Note:
Matrix Inversion method can be applied only when the coefficient matrix is a square matrix and non-singular. The different types of inversion methods are Gauss-Jordan Elimination, Classical Adjoint method, Partition method.