Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (A)

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