Question
Find the inverse of $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$
✓
Answer
As $|A|=\left|\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right|=-2$
$\therefore \quad| A | \neq 0 \quad \therefore A ^{-1}$ exists.
Let $\quad AA ^{-1}= I ($Here we can use only row transformation$)$
Using $R_2 \rightarrow R_2-3 R_1$
$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad A ^{-1}=\left[\begin{array}{ll}
1 & 0 \\ 0 & 1\end{array}\right]$ becomes
$\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right]$
Using $-\frac{1}{2} R_2$ we get
$\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ \frac{3}{2} & -\frac{1}{2}
\end{array}\right]$
Using $R_1 \rightarrow R_1-2 R_2$
We get $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\therefore \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right] \quad ($Verify the answer.$)$
$\therefore \quad| A | \neq 0 \quad \therefore A ^{-1}$ exists.
Let $\quad AA ^{-1}= I ($Here we can use only row transformation$)$
Using $R_2 \rightarrow R_2-3 R_1$
$\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right] \quad A ^{-1}=\left[\begin{array}{ll}
1 & 0 \\ 0 & 1\end{array}\right]$ becomes
$\left[\begin{array}{cc} 1 & 2 \\ 0 & -2 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ -3 & 1 \end{array}\right]$
Using $-\frac{1}{2} R_2$ we get
$\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right] A ^{-1}=\left[\begin{array}{cc} 1 & 0 \\ \frac{3}{2} & -\frac{1}{2}
\end{array}\right]$
Using $R_1 \rightarrow R_1-2 R_2$
We get $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$
$\therefore \quad A ^{-1}=\left[\begin{array}{cc}-2 & 1 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right] \quad ($Verify the answer.$)$
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