Question

If $B^n-A=I$ and $A=\begin{bmatrix}26&26&18\\ 25&37&17\\ 52&39&50\end{bmatrix}, B=\begin{bmatrix}1&4&2\\ 3&5&1\\ 7&1&6\end{bmatrix},$ then $n =$

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

$\because B^{n}-A=I$
$\therefore B^{n}=I+A$
$B^{n}=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}+\begin{bmatrix}26&26&18\\ 25&37&17\\ 52&39&50\end{bmatrix}$
$B^{n}=\begin{bmatrix}27&26&18\\ 25&38&17\\ 52&39&51\end{bmatrix}$
or $\begin{bmatrix}1&4&2\\ 3&5&1\\ 7&1&6\end{bmatrix}^{^{^n}}=\begin{bmatrix}27&26&18\\ 25&38&17\\ 52&39&51\end{bmatrix}\,...\left(i\right)$
$\therefore n \ne1$
Now put $n = 2,$ then
$B^{2}=\begin{bmatrix}1&4&2\\ 3&5&1\\ 7&1&6\end{bmatrix}^{^2}=\begin{bmatrix}1&4&2\\ 3&5&1\\ 7&1&6\end{bmatrix}$$\begin{bmatrix}1&4&2\\ 3&5&1\\ 7&1&6\end{bmatrix}$
$=\begin{bmatrix}1+12+14&4+20+2&2+4+12\\ 3+15+7&12+25+1&6+5+6\\ 7+3+42&28+5+6&14+1+36\end{bmatrix}$
$=\begin{bmatrix}27&26&18\\ 25&38&17\\ 52&39&51\end{bmatrix}$
Which is equal to R.H.S. of eq. (i).
$\therefore n=2$