Question

If \[A=[\begin{matrix}
   \text{ }\!\!~\!\!\text{ }a & b\text{ }\!\!~\!\!\text{ } \\
\end{matrix}]\], $B=[\begin{matrix}
   -b & -a\text{ }\!\!~\!\!\text{ } \\
\end{matrix}]$ and $c=\left[ \begin{matrix}
   \text{ }\!\!~\!\!\text{ }a \\
   \text{-}a\text{ }\!\!~\!\!\text{ } \\
\end{matrix} \right]$, then correct statement is
A $A = - B$
B $A + B = A - B$
C $AC = BC$
D $CA = CB$

Answer 1

Hint:To solve this question we will take each of the options and then substitute the values of the given matrices and check if they are correct or not. In the first option we will take matrix B and multiply it with $-1$ and then compare it with matrix B and see if it is correct or not. In the second option we will take the sum of the matrices of A and B and the difference of matrices A and B and then compare both the results to check if it is correct or not. In the third option we will find the product of matrices A and C and product of matrices B and C. Then we will check whether AC is equal to BC or not to get the correct option. To check the fourth option we will use the property of matrices which states that commutative property does not apply on matrices.

Complete step by step Solution:
We will take first option $A = - B$,
$\begin{align}
  & -B=-[\begin{matrix}
   b & a \\
\end{matrix}] \\
 & -B=[-\begin{matrix}
   b & -a \\
\end{matrix}] \\
\end{align}$

$\begin{align}
  & A=-B \\
 & [\begin{matrix}
   a & b \\
\end{matrix}]\ne [\begin{matrix}
   b & a \\
\end{matrix}]
\end{align}$
Hence the first option is incorrect.
We will now take the second option $A + B = A - B$.
We will first find the sum of the matrices A and B.
$\begin{align}
  & A+B=[\begin{matrix}
   a & b \\
\end{matrix}]+[\begin{matrix}
   -b & -a \\
\end{matrix}] \\
 & =[\begin{matrix}
   a-b & b \\
\end{matrix}-a]
\end{align}$
Now the difference of matrices A and B.
$\begin{align}
  & A-B=[\begin{matrix}
   a & b \\
\end{matrix}]-[\begin{matrix}
   -b & -a \\
\end{matrix}] \\
 & =[\begin{matrix}
   a+b & b \\
\end{matrix}+a]
\end{align}$
Comparing sum and difference of matrices A and B,
$\begin{align}
  & A+B=A-B \\
 & [\begin{matrix}
   a-b & b\\
\end{matrix}-a]\ne [\begin{matrix}
   a+b & b \\
\end{matrix}+a]
\end{align}$
Hence this option is also incorrect.
We will now take the third option $AC = BC$.
$AC=[\begin{matrix}
   \text{ }~\text{ }a & b\text{ }~\text{ } \\
\end{matrix}]\left[ \begin{matrix}
   \text{ }~\text{ }a \\
   -a\text{ }~\text{ } \\
\end{matrix} \right]$After multiplying we will get
$AC = \left[ {a.a + b( - a)} \right]$
After solving we get
$AC = \left[ {{a^2} - ab} \right]$
Now we will find $BC$,
$BC=[\begin{matrix}
   -b & -a \\
\end{matrix}]\left[ \begin{matrix}
   \text{ }~\text{ }a \\
   -a\text{ }~\text{ } \\
\end{matrix} \right]$After multiplying we will get
$BC = \left[ {( - b)a + ( - a)( - a)} \right]$
After solving we get
$BC = \left[ { - ba + {a^2}} \right]$
$BC = \left[ {{a^2} - ab} \right]$
Therefore, $AC = BC$.
Hence, option C is correct.
Because we know that multiplication in matrices is not commutative so as $AC = BC$ then $CA\ne CB$. Hence the fourth option is also incorrect.

Therefore, the correct option is (C).

Note:While multiplying the matrices we have to keep in mind the order of the matrices. In finding the product of $AC$ there are different orders of both the matrix as $A$ has order \[1\times 2\] and $C$ has order \[2\times 1\] . And in $BC$ the order of the matrix of $B$ is \[1\times 2\] and of $C$ is \[2\times 1\]. So we have to do the multiplication accordingly.