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

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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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.
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