Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
Solve the following by inversion method $2x + y = 5, 3x + 5y = −3$
✓
Answer
Matrix form of the given system of equations is
$\left[\begin{array}{ll}2 & 1 \\3 & 5\end{array}\right]\left[\begin{array}{l}x \\y\end{array}\right]=\left[\begin{array}{c}5 \\-3\end{array}\right]$
This is of the form $A X=B$,
where $A =\left[\begin{array}{ll}2 & 1 \\ 3 & 5\end{array}\right], X =\left[\begin{array}{l}x \\ y\end{array}\right]$ and
$B =\left[\begin{array}{c}5 \\ -3\end{array}\right]$
To determine $X$, we have to find $A^{-1}$
$\begin{array}{l}|A|=\left[\begin{array}{ll}2 & 1 \\3 & 5\end{array}\right] \\=10-3 \\=7 \neq 0\end{array}$
$\therefore A ^{-1}$ exists.
Consider $AA ^{-1}= I$
$\therefore\left[\begin{array}{ll}2 & 1 \\3 & 5\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Applying $R_2 \rightarrow 2 R_2-3 R_1$, we get
$\left[\begin{array}{ll}2 & 1 \\0 & 7\end{array}\right] A^{-1}=\left[\begin{array}{cc}1 & 0 \\-3 & 2\end{array}\right]$
Applying $R_1 \rightarrow 7 R_1-R_2$, we get
$\left[\begin{array}{cc}14 & 0 \\0 & 7\end{array}\right] A^{-1}=\left[\begin{array}{cc}10 & -2 \\-3 & 2\end{array}\right]$
Applying $R _1 \rightarrow\left(\frac{1}{14}\right) R _1$
and $R _2 \rightarrow\left(\frac{1}{7}\right) R _2$, we get
${\left[\begin{array}{ll}1 & 0 \\0 & 1 \end{array}\right] A^{-1}=\left[\begin{array}{cc}\frac{10}{14} & \frac{-2}{14} \\\frac{-3}{7} & \frac{2}{7}\end{array}\right]} $
$\therefore A^{-1}=\frac{1}{7}\left[\begin{array}{cc}5 & -1 \\-3 & 2\end{array}\right]$
Pre$-$multiplying $A X=B$ by $A^{-1}$, we get
$A^{-1}(A X)=A^{-1} B$
$\therefore\left( A ^{-1} A \right) X = A ^{-1} B$
$\therefore IX = A ^{-1} B$
$\therefore X = A ^{-1} B$
$\therefore X =\frac{1}{7}\left[\begin{array}{cc}5 & -1 \\-3 & 2\end{array}\right]\left[\begin{array}{c}5 \\-3\end{array}\right] $
$\therefore\left[\begin{array}{c}x \\y\end{array}\right]=\frac{1}{7}\left[\begin{array}{c}25+3 \\-15-6\end{array}\right] $
$=\frac{1}{7}\left[\begin{array}{c}28 \\-21\end{array}\right] $
$=\left[\begin{array}{c}4 \\-3\end{array}\right]$
$\therefore$ By equality of matrices, we get
$x=4, y=-3$
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