Question
If $A=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$ and $A^2=I$; Find $a$ and $b$
✓
Answer
$A=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$
$A^2=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$
$=\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]$
It is given that $A^2=I$
$\therefore\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Comparing the corresponding elements we get
1 + a = 1
Therefore a = 0
-1 + b = 0
Therefore b = 1
$A^2=\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]\left[\begin{array}{cc}-1 & 1 \\ a & b\end{array}\right]$
$=\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]$
It is given that $A^2=I$
$\therefore\left[\begin{array}{cc}1+a & -1+b \\ -a+a b & a+b^2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Comparing the corresponding elements we get
1 + a = 1
Therefore a = 0
-1 + b = 0
Therefore b = 1
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