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Matrices (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Matrices (p-1)3 Marks
Question
Find inverse of the following matrices (if they exist) by elementary transformation : $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$
✓

Answer

Let $A=\left(\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right)$
Then $|A|=\left|\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right|=3-(-2)=5 \neq 0$
$\therefore A ^{-1}$ exists.
We write, $AA ^{-1}= I$
$
\therefore\left(\begin{array}{cc}
1 & -1 \\
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}{cc}
1 & -1 \\
0 & 5
\end{array}\right] A^{-1}=\left(\begin{array}{cc}
1 & 0 \\
-2 & 1
\end{array}\right)
$
By $\left(\frac{1}{5}\right) R_2$, we get
$
\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right] A^{-1}=\left[\begin{array}{rr}
1 & 0 \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]
$
By $R_1+R_2$, we get
$
\begin{aligned}
& {\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] A ^{-1}=\left[\begin{array}{rr}
3 & \frac{1}{5} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]} \\
& \therefore A ^{-1}=\left[\begin{array}{rr}
\frac{3}{5} & \frac{1}{5} \\
-\frac{2}{5} & \frac{1}{5}
\end{array}\right]
\end{aligned}
$

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