Question
If $A=\left[\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right]$, and $A B=B A=1$, then find the matrix $B$
✓
Answer
$A B=B A=1$
$\Rightarrow B=A^{-1}$
$|A|=\left|\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right|$
$=6-5$
$=1$
$\therefore A_{11}=(-1)^{1+1} M_{11}=M_{11}=2$
$A_{12}=(-1)^{1+2} M_{12}=-M_{12}=-5$
$A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-1$
$A_{22}=(-1)^{2+2} M_{22}=M_{22}=3$
$\therefore \operatorname{adj}(A)=\left[\begin{array}{cc}2 & -5 \\ -1 & 3\end{array}\right]^T$
$=\left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right]$
$\begin{array}{l}\therefore B=A^{-1}=\frac{1}{|A|} \operatorname{adj}(A) \end{array}$
$ \therefore B=\left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right]$
$\Rightarrow B=A^{-1}$
$|A|=\left|\begin{array}{ll}3 & 1 \\ 5 & 2\end{array}\right|$
$=6-5$
$=1$
$\therefore A_{11}=(-1)^{1+1} M_{11}=M_{11}=2$
$A_{12}=(-1)^{1+2} M_{12}=-M_{12}=-5$
$A_{21}=(-1)^{2+1} M_{21}=-M_{21}=-1$
$A_{22}=(-1)^{2+2} M_{22}=M_{22}=3$
$\therefore \operatorname{adj}(A)=\left[\begin{array}{cc}2 & -5 \\ -1 & 3\end{array}\right]^T$
$=\left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right]$
$\begin{array}{l}\therefore B=A^{-1}=\frac{1}{|A|} \operatorname{adj}(A) \end{array}$
$ \therefore B=\left[\begin{array}{cc}2 & -1 \\ -5 & 3\end{array}\right]$
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