Question

if A=[102021203], prove that A3−6A2+7A+2I=0

Answer 1

$\textbf{Given:}$

$A=\left(\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right)$

$\textbf{To prove:}$

$A^3-6\,A^2+7A+2I=0$

$\textbf{Solution:}$

$\text{Consider,}$

$A^2=\left(\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right)\left(\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right)$

$A^2=\left(\begin{array}{ccc}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9\end{array}\right)$

$A^2=\left(\begin{array}{ccc}5&0&8\\2&4&5\\8&0&13\end{array}\right)$

$A^3=A^2{\times}A$

$A^3=\left(\begin{array}{ccc}5&0&8\\2&4&5\\8&0&13\end{array}\right)\left(\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right)$

$A^3=\left(\begin{array}{ccc}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39\end{array}\right)$

$A^3=\left(\begin{array}{ccc}21&0&34\\12&8&23\\34&0&55\end{array}\right)$

$\text{Now,}$

$A^3-6\,A^2+7A+2I=0$

$=\left(\begin{array}{ccc}21&0&34\\12&8&23\\34&0&55\end{array}\right)-6\left(\begin{array}{ccc}5&0&8\\2&4&5\\8&0&13\end{array}\right)+7\left(\begin{array}{ccc}1&0&2\\0&2&1\\2&0&3\end{array}\right)+2\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

$=\left(\begin{array}{ccc}21&0&34\\12&8&23\\34&0&55\end{array}\right)-\left(\begin{array}{ccc}30&0&48\\12&24&30\\48&0&78\end{array}\right)+\left(\begin{array}{ccc}7&0&14\\0&14&7\\14&0&21\end{array}\right)+\left(\begin{array}{ccc}2&0&0\\0&2&0\\0&0&2\end{array}\right)$

$=\left(\begin{array}{ccc}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{array}\right)$

$=\left(\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right)$

$\implies\bf\,A^3-6\,A^2+7\,A+2\,I=0$