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

Last updated date: 19th Sep 2024
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Hint:To find the value of $x$ and $y$ we will first form a matrix from the two equation and then find the determinant of the matrix A and then we will find the inverse of matrix A and form product with a $2\times 1$ matrix of constant value of the equation given as:
$X={{A}^{-1}}B$

Complete step by step solution:
The two equation given are $2x-3y+6=0$ and $6x+y+8=0$, and to form the matrix A we will form the matrix A as $\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|$ which is equal to $\left| \begin{matrix} 2 & -3 \\ 6 & 1 \\ \end{matrix} \right|$ and to form the matrix B we will make a $2\times 1$ matrix of constant value of
$\left| \begin{matrix} m \\ n \\ \end{matrix} \right|$ as $\left| \begin{matrix} -6 \\ -8 \\ \end{matrix} \right|$.
With the matrix of X as $\left| \begin{matrix} x \\ y \\ \end{matrix} \right|$ we will form a matrix equation of:
$\left| \begin{matrix} x \\ y \\ \end{matrix} \right|={{A}^{-1}}\left| \begin{matrix} m \\ n \\ \end{matrix} \right|$
Now forming the inverse of the matrix A, we will get the inverse of matrix A as:
${{A}^{-1}}=\dfrac{1}{\left| A \right|}\left| \begin{matrix} a & b \\ c & d \\ \end{matrix} \right|$
The value of $\left| A \right|$ is the determinant which is given as:
$\Rightarrow \left| A \right|=\left( 2\times 1-\left( -3\times 6 \right) \right)$
$\Rightarrow \left| A \right|=20$
Now with the determinant value found we will find the value of inverse matrix of A as:
${{A}^{-1}}=\dfrac{1}{20}\left| \begin{matrix} 1 & 3 \\ -6 & 2 \\ \end{matrix} \right|$ (The inverse of $\left| \begin{matrix} 2 & -3 \\ 6 & 1 \\ \end{matrix} \right|$ is $\left| \begin{matrix} 1 & 3 \\ -6 & 2 \\ \end{matrix} \right|$ by interchanging the original matrix as $\left| \begin{matrix} d & -b \\ -c & a \\ \end{matrix} \right|$)
$\Rightarrow {{A}^{-1}}=\dfrac{1}{20}\left| \begin{matrix} 1 & 3 \\ -6 & 2 \\ \end{matrix} \right|$
Placing the inverse value in $\left| \begin{matrix} x \\ y \\ \end{matrix} \right|={{A}^{-1}}\left| \begin{matrix} m \\ n \\ \end{matrix} \right|$ , we get:
$\Rightarrow \left| \begin{matrix} x \\ y \\ \end{matrix} \right|=\dfrac{1}{20}\left| \begin{matrix} 1 & 3 \\ -6 & 2 \\ \end{matrix} \right|\left| \begin{matrix} -6 \\ -8 \\ \end{matrix} \right|$
$\Rightarrow \left| \begin{matrix} x \\ y \\ \end{matrix} \right|=\dfrac{1}{20}\left| \begin{matrix} 1\times -6+3\times -8 \\ -6\times -6+2\times 8 \\ \end{matrix} \right|$
$\Rightarrow \left| \begin{matrix} x \\ y \\ \end{matrix} \right|=\left| \begin{matrix} \dfrac{1\times -6+3\times -8}{20} \\ \dfrac{-6\times -6+2\times 8}{20} \\ \end{matrix} \right|$
$\Rightarrow \left| \begin{matrix} x \\ y \\ \end{matrix} \right|=\left| \begin{matrix} \dfrac{-30}{20} \\ \dfrac{52}{20} \\ \end{matrix} \right|$
Therefore, the value of $x=\dfrac{-3}{2}$and $y=\dfrac{13}{5}$

Note: The matrix inversion method can only work on a square matrix. We will also solve these equations by elimination method and substitution method.