Question

Consider three matrices $A=\left[\begin{array}{cc}2& 1\\ 4& 1\end{array}\right]B=\left[\begin{array}{cc}3& 4\\ 2& 3\end{array}\right]$ and $C=\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right]$.
The value of the sum $tr\left(A\right)+tr\left(\frac{ABC}{2}\right)+tr\left(\frac{A({BC)}^2}{4}\right)+tr\left(\frac{A({BC)}^3}{8}\right)+.............+\infty$ is
($tr\left(A\right)$ denotes trace of a matrix $A$)

• A. $6$
• B. $9$
• C. $12$
• D. none of these

Answer 1

The correct option is A $6$

$A=\left[\begin{array}{cc}2& 1\\ 4& 1\end{array}\right]$
So,$tr\left(A\right)=2+1=3$
Now, $BC=\left[\begin{array}{cc}3& 4\\ 2& 3\end{array}\right]\left[\begin{array}{cc}3& -4\\ -2& 3\end{array}\right]$
$BC=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=I$
Now, $ABC=A.I=A$

So, $\frac{ABC}{2}=\frac{1}{2}A$

$\Rightarrow tr\left(\frac{ABC}{2}\right)=\frac{3}{2}$
Now, $A(BC{)}^2=A.I=A$

So, $\frac{A(BC{)}^2}{4}=\frac{1}{4}A$

$\Rightarrow tr\left(\frac{A(BC{)}^2}{4}\right)=\frac{3}{4}$
Hence, $tr\left(A\right)+tr\left(\frac{ABC}{2}\right)+tr\left(\frac{A({BC)}^2}{4}\right)+tr\left(\frac{A({BC)}^3}{8}\right)+.............+\infty$
$=3+\frac{3}{2}+\frac{3}{4}+......$
which is an infinite G.P.
$S=\frac{3}{1-\frac{1}{2}}=6$