Question

Let matrix $A=\left[\begin{array}{ccc}1& 1& m\\ p& a& n\\ q& r& 1\end{array}\right],$ $a,m,n,p,q,r,\in R$ be an idempotent matrix. Then the value of $|A|+\frac{|A|}{tr\left(A\right)}+\frac{|A{|}^2}{\left(tr\right(A^2){)}^2}+\frac{|A^3|}{\left(tr\right(A^3){)}^3}+\cdots +\infty \text{terms,}$ $($where $|A|\ne 0$ and $tr\left(A\right)\ne 0)$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{4}{3}$
• C. $2$
• D. $1$

Answer 1

The correct option is A $\frac{3}{2}$

$A^2=A$ and $|A|\ne 0$
$\therefore A^{-1}A\cdot A=A^{-1}A$
$\Rightarrow A=I$
$\therefore |A|=1$ and $tr\left(A\right)=3$
Hence, the given series is an infinite G.P.
Here, $a=1,r=\frac{1}{3},S_{\infty }=\frac{a}{1-r}$
$\therefore$ Given sum $=1+\frac{1}{3}+\frac{1}{3^2}+\cdots =\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$