Question

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

Answer 1

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

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.