Question
Find $A^{-1}$ using column transformations:$A=\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right]$
✓
Answer
We know that $A A^{-1}=1$
$\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Applying $C _1 \rightarrow 2 C _1+ C _2$, we get
$\left[\begin{array}{cc}1 & -3 \\0 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 0 \\1 & 1\end{array}\right]$
Applying $C_2 \rightarrow C_2+3 C_1$, we get
$\left[\begin{array}{ll}1 & 0 \\0 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 6 \\1 & 4\end{array}\right]$
Applying $C_2 \rightarrow\left(\frac{1}{2}\right) C_2$, we get
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A ^{-1}=\left[\begin{array}{ll}2 & 3 \\1 & 2\end{array}\right]} \end{array}$
$\therefore A ^{-1}=\left[\begin{array}{ll}2 & 3 \\1 & 2\end{array}\right]$
$\left[\begin{array}{cc}2 & -3 \\-1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]$
Applying $C _1 \rightarrow 2 C _1+ C _2$, we get
$\left[\begin{array}{cc}1 & -3 \\0 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 0 \\1 & 1\end{array}\right]$
Applying $C_2 \rightarrow C_2+3 C_1$, we get
$\left[\begin{array}{ll}1 & 0 \\0 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 6 \\1 & 4\end{array}\right]$
Applying $C_2 \rightarrow\left(\frac{1}{2}\right) C_2$, we get
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right] A ^{-1}=\left[\begin{array}{ll}2 & 3 \\1 & 2\end{array}\right]} \end{array}$
$\therefore A ^{-1}=\left[\begin{array}{ll}2 & 3 \\1 & 2\end{array}\right]$
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