Question

If $A=\left[\begin{array}{cc}a& b\end{array}\right],B=\left[\begin{array}{cc}-b& -a\end{array}\right]$ and $C=\left[\begin{array}{c}a\\ -a\end{array}\right]$, then the correct statement is

• A. $A=-B$
• B. $A+B=A-B$
• C. $AC=BC$
• D. $CA=CB$

Answer 1

The correct option is C $AC=BC$

Given $A=\left[\begin{array}{cc}a& b\end{array}\right],B=\left[\begin{array}{cc}-b& -a\end{array}\right]$ and $C=\left[\begin{array}{c}a\\ -a\end{array}\right]$
We will check by options.
Clearly , $A\ne -B$ as their corresponding elements are different only.
$A+B=\left[\begin{array}{cc}a-b& b-a\end{array}\right]$
$A-B=\left[\begin{array}{cc}a+b& b+a\end{array}\right]$
So, $A+B\ne A-B$
Now,$AC=\left[\begin{array}{cc}a& b\end{array}\right]\left[\begin{array}{c}a\\ -a\end{array}\right]$
$\Rightarrow AC=\left[\begin{array}{c}{a}^2-ab\end{array}\right]$
$BC=\left[\begin{array}{cc}-b& -a\end{array}\right]\left[\begin{array}{c}a\\ -a\end{array}\right]$
$\Rightarrow BC=\left[\begin{array}{c}{a}^2-ab\end{array}\right]$
Hence, $AC=BC$
Option $C$ is correct
Now, $CA=\left[\begin{array}{c}a\\ -a\end{array}\right]\left[\begin{array}{cc}a& b\end{array}\right]$
$\Rightarrow CA=\left[\begin{array}{cc}{a}^2& ab\\ -a^2& -ab\end{array}\right]$
$CB=\left[\begin{array}{c}a\\ -a\end{array}\right]\left[\begin{array}{cc}-b& -a\end{array}\right]$
$\Rightarrow CB=\left[\begin{array}{cc}-ab& -a^2\\ ab& a^2\end{array}\right]$
Hence, $CA\ne CB$