Question

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

Answer 1

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.