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Matrics — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsMatrics4 Marks
Question
Solve the following equations by inversion method : $x + 2y = 2, 2x + 3y = 3$
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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
$A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}x \\ y\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}2 \\ 3\end{array}\right]$
Let us find $A^{-1}$.
$|A|=\left|\begin{array}{ll}1 & 2 \\2 & 3\end{array}\right|=3-4=-1 \neq 0$
$\therefore \mathrm{A}^{-1}$ exists.
Consider $\mathrm{AA}^{-1}=\mathrm{I}$
$\therefore\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
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,
$\begin{array}{l}\quad\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right] \\ \therefore A^{-1}=\left[\begin{array}{rr}-3 & 2 \\ 2 & -1\end{array}\right]\end{array}$
Now, premultiply $AX = B$ by $A^{-1}$, we get,
$A^{-1}(AX) = A^{-1}B$
$\therefore (A^{-1}A)X = A^{-1}B$
$\therefore IX = A^{-1}B$
$\therefore \mathrm{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.

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