Question

Let $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ and $B\left[\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right]$. If $10A^{10}+adj\left(A^{10}\right)=B$, then $b_1+b_2+b_3+b_4$ is equal to

• A. $91$
• B. $92$
• C. $111$
• D. $112$

Answer 1

The correct option is D $112$

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ $B=\left[\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 1+1\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$

$A^3=A^2.A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 1+2\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$

$A^4=A^3.A=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& 1+3\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]$

similarly $A^{10}=\left[\begin{array}{cc}1& 10\\ 0& 1\end{array}\right]$

Adj $A=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]=\left[\begin{array}{cc}1& -10\\ 0& 1\end{array}\right]$

$10A^{10}+adj\left(A^{10}\right)=10\left[\begin{array}{cc}1& 10\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}1& -10\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}10& 100\\ 0& 10\end{array}\right]+\left[\begin{array}{cc}1& -10\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}11& 90\\ 0& 11\end{array}\right]$

$10A^{10}+adj{A}^{10}=\left[\begin{array}{cc}11& 90\\ 0& 11\end{array}\right]=B=\left[\begin{array}{cc}{b}_1& b_2\\ b_3& b_4\end{array}\right]$

$b_1+b_2+b_3+b_4=11+90+0+11=112$