Solve the following equations by the method of inversion : x + 2y…

Question

Solve the following equations by the method of inversion : $x + 2y = 2, 2x + 3y = 3$

✓

Answer

The given equations can be written in the matrix form as:
$ \left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left(\begin{array}{l} 2 \\ 3 \end{array}\right]$
This is of the form $AX = B$, where
$ A =\left(\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right), X =\left[\begin{array}{l} x \\ y \end{array}\right) \text { and } B =\left(\begin{array}{l} 2 \\ 3 \end{array}\right]$
Let us find $A ^{-1}$.
$| A |=\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|=3-4=-1 \neq 0$
$\therefore A ^{-1}$ exists.
We write $AA ^{-1}= I$
$\therefore\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] A ^{-1}=\left[\begin{array}{ll} 1 & 0 \\
0 & 1 \end{array}\right]$
By $R_2-2 R_1$, we get
$\left[\begin{array}{rr} 1 & 2 \\ 0 & -1 \end{array}\right] A ^{-1}=\left(\begin{array}{rr} 1 & 0 \\ -2 & 1
\end{array}\right]$
By $(-1) R_2$, we get
$\left(\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right) A^{-1}=\left(\begin{array}{rr} 1 & 0 \\ 2 & -1 \end{array}\right)$
By $R_1-2 R_2$, we get
$ \begin{aligned} & \quad\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{rr} -3 & 2 \\ 2 & -1 \end{array}\right] \\ & \therefore A ^{-1}=\left(\begin{array}{rr} -3 & 2 \\ 2 & -1 \end{array}\right]
\end{aligned}$
Now, premultiply $A X=B^{-1}$ by $A^{-1}$, we get
$ \begin{aligned} & A ^{-1}( AX )= A ^{-1} B \\ \therefore & \left( A ^{-1} A \right) X = A ^{-1} B \\ \therefore & IX = A ^{-1} B \end{aligned} $
$ \therefore X=\left[\begin{array}{rr} -3 & 2 \\ 2 & -1 \end{array}\right]\left[\begin{array}{l}
2 \\ 3\end{array}\right] $
$ \therefore\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{r} -6+6 \\ 4-3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 1 \end{array}\right]$
By equality of matrices, $x=0, y=1$ is the required solution.