Question

solve this question properly on paper ..
according to 12th class​

Answer 1

Answer:

hence proved

Step-by-step explanation:

$A=\left[\begin{array}{ccc}1&2&3\\3&-2&1\\4&2&1\end{array}\right]$

$A^{2} =\left[\begin{array}{ccc}1&2&3\\3&-2&1\\4&2&1\end{array}\right] \left[\begin{array}{ccc}1&2&3\\3&-2&1\\4&2&1\end{array}\right] =\left[\begin{array}{ccc}19&4&8\\1&12&6\\14&2&15\end{array}\right]$

$A^{3} = \left[\begin{array}{ccc}19&4&8\\1&12&6\\14&2&15\end{array}\right] \left[\begin{array}{ccc}1&2&3\\3&-2&1\\4&2&1\end{array}\right] = \left[\begin{array}{ccc}63&46&69\\61&2&21\\80&54&59\end{array}\right]$

$23A= \left[\begin{array}{ccc}23&46&69\\69&-46&23\\92&46&23\end{array}\right]$

$40I= \left[\begin{array}{ccc}40&0&0\\0&40&0\\0&0&40\end{array}\right]$

$A^{3} -23A-40I= \left[\begin{array}{ccc}63&46&69\\1&12&6\\14&2&15\end{array}\right] - \left[\begin{array}{ccc}23&46&69\\69&-46&23\\92&46&23\end{array}\right] - \left[\begin{array}{ccc}40&0&0\\0&40&0\\0&0&40\end{array}\right]$

$A^{3}-23A-40I= \left[\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right]$