Question
If $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ show that $A^2-(a+d) A=(b c-a d) I_2$
✓
Answer
$\begin{aligned} & A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right], I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ & A^2=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \times\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ & =\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & b c+d^2\end{array}\right] \\ & \text { L.H.S. }=A^2-(a+d) A \\ & =\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+c d & b c+d^2\end{array}\right]-(a+d)\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \\ & =\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+c d & b c+d^2\end{array}\right]-\left[\begin{array}{ll}a^2+a d & a b+b d \\ a c+c d & a d+d^2\end{array}\right] \\ & =\left[\begin{array}{cc} bc - ad & 0 \\ 0 & bc - ad \end{array}\right] \\ & =( bc - ad )\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\ & =(b c-a d)I \\ & \text { L.H.S. = R.H.S. } \\ & \end{aligned}$
$A^2-(a+d) A=(b c-a d) l_2$
$A^2-(a+d) A=(b c-a d) l_2$
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