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 $A^2 - (a + d) A = (bc - ad) I.$
✓
Answer
Here $A^2 - (a + d) A$
$\begin{array}{l}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]-(a+d)\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\end{array}$
$=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & c b+d^2\end{array}\right]-\left[\begin{array}{ll}a^2+a d & a b+b d \\ a c+d c & a d+d^2\end{array}\right] $
$ =\left[\begin{array}{cc}b c-a d & 0 \\ 0 & b c-a d\end{array}\right]$
$=(b c-a d)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$= (bc - ad) I.$
Hence proved
$\begin{array}{l}=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]-(a+d)\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\end{array}$
$=\left[\begin{array}{ll}a^2+b c & a b+b d \\ a c+d c & c b+d^2\end{array}\right]-\left[\begin{array}{ll}a^2+a d & a b+b d \\ a c+d c & a d+d^2\end{array}\right] $
$ =\left[\begin{array}{cc}b c-a d & 0 \\ 0 & b c-a d\end{array}\right]$
$=(b c-a d)\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$= (bc - ad) I.$
Hence proved
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