Question

Let $A_n=[a_{ij}{]}_{n\times n}$ be a $(n\times n)$ determinant with the following conditions
$a_{ij}=\left\{\begin{array}{cc}9,& i=j\\ 3,& |i-j|=1\\ 0,& \text{other wise}\end{array}\right\}$ then

• A. $A_3=567$
• B. $A_6+9A_4=9A_5$
• C. $A_6+9A_4=3A_5$
• D. $A_{12}+8A_{10}=81A_{11}$

Answer 1

Answer: (A), (B)

The correct options are
A $A_3=567$
B $A_6+9A_4=9A_5$

$A_n={\left|\begin{array}{ccccccc}9& 3& 0& 0& 0& 0& ----\\ 3& 9& 3& 0& 0& 0& ----\\ 0& 3& 9& 3& 0& 0& ----\\ 0& 0& 3& 9& 3& 0& ---\end{array}\right|}_{(n\times n)}$
$A_n=9\times {\left|\begin{array}{ccccc}9& 3& 0& 0& 0----\\ 3& 9& 3& & ----\end{array}\right|}_{(n-1)}-3\times {\left|\begin{array}{ccccccc}3& 3& 0& 0& 0& 0& -\\ 0& 9& 3& -& -& -& -\end{array}\right|}_{(n-1)}$
$=9\times A_{n-1}-3\times 3\times \left|\begin{array}{ccccc}1& 1& 0& ---& \\ 0& 9& 3& ---& \\ 0& & & & \\ 0& & & & \end{array}\right|$
$A_n=9A_{n-1}-9A_{n-2}{A}_1=9,A_2=72$
$A_3=9(72-9)=729$
$\therefore A_6=9A_5-9A_4\Rightarrow A_6+9A_4=9A_5$
$Also{A}_{12}=9A_{11}-9A_{10}\Rightarrow A_{12}+9A_{10}=9A_{11}$