Question

if A=[a11=2 a12=3 a21=4 a22=5] show that A^2-7A-21=0​

Answer 1

Given

$\tt\to A = \left[\begin{array}{ccc}2&3\\4&5\end{array}\right]$

Show That

$\tt\to A^2-7A -2I=0$

Where I is

$\tt\to I = \left[\begin{array}{ccc}\tt 1&\tt 0\\\tt 0&\tt 1\end{array}\right]$

We Know that

$\tt \to AB= \left[\begin{array}{ccc}\tt a& \tt b\\\tt c&\tt d\end{array}\right]\left[\begin{array}{ccc}\tt e& \tt f\\\tt g&\tt h\end{array}\right]=\left[\begin{array}{ccc}\tt ae+bg & \tt af+bh\\\tt ce+dg&\tt cf+dh\end{array}\right]$

We get

$\tt\to AA=\left[\begin{array}{ccc}\tt 2& \tt 3\\\tt 4&\tt 5\end{array}\right]\left[\begin{array}{ccc}\tt 2& \tt 3\\\tt 4&\tt 5\end{array}\right]=\left[\begin{array}{ccc}\tt 4+12& \tt 6+15\\\tt 8+20&\tt 12+25\end{array}\right]$

$\tt\to A^2=\left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]$

Now We have to Find

$\tt\to A^2-7A -2I$

Put the value

$\tt\to \left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]-7\left[\begin{array}{ccc}2&3\\4&5\end{array}\right]-2 \left[\begin{array}{ccc}\tt 1&\tt 0\\\tt 0&\tt 1\end{array}\right]$

$\tt\to \left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]-\left[\begin{array}{ccc}14&21\\28&35\end{array}\right]-\left[\begin{array}{ccc}\tt 2&\tt 0\\\tt 0&\tt 2\end{array}\right]$

$\tt\to \left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]-\bigg(\left[\begin{array}{ccc}14&21\\28&35\end{array}\right]+\left[\begin{array}{ccc}\tt 2&\tt 0\\\tt 0&\tt 2\end{array}\right]\bigg)$

$\tt\to \left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]-\bigg(\left[\begin{array}{ccc}14+2&21+0\\28+0&35+2\end{array}\right]\bigg)$

$\tt\to \left[\begin{array}{ccc}\tt 16& \tt 21\\\tt 28&\tt 37\end{array}\right]-\left[\begin{array}{ccc}16&21\\28&37\end{array}\right]$

$\tt\to \left[\begin{array}{ccc}\tt 16-16& \tt 21-21\\\tt 28-28&\tt 37-37\end{array}\right]$

$\tt\to \left[\begin{array}{ccc}\tt 0& \tt 0\\\tt 0&\tt 0\end{array}\right]=0$

Hence Proved