Question

If $A$ and $B$ are two square matrices of order $n$ and $A$ and $B$ commute then for any real number $K$, then

• A. $A-KI,B-KI$ commute.
• B. $A-KI,B-KI$ are equal.
• C. $A-KI,B-KI$ do not commute.
• D. $A+KI,B-KI$ commute.

Answer 1

The correct option is A $A-KI,B-KI$ commute.

$AB=BA$ (GIven)
Consider $(A-KI)(B-KI)$

where, $K$ is any real number
$(A-KI)(B-KI)=AB-K(A+B)+K^2$
Interchanging $A$ and $B$, we get

$(B-KI)(A-KI)=BA-K(B+A)+K^2$
Since $A,B$ commute with each other

$\therefore (B-KI)(A-KI)=BA-K(B+A)+K^2=AB-K(A+B)+K^2=(A-KI)(B-KI)$

Hence, option A is true and option C is false.

We cannot say anything about equality of $A-KI$ and $B-KI$, since it is not given that $A$ and $B$ are equal or not.

Hence, option B is false.

Consider $(A+KI)(B-KI)$

$(A+KI)(B-KI)=AB-K(A-B)-K^2$

Since $A,B$ commute with each other,

$\therefore (B-KI)(A+KI)=BA-K(B+A)+K^2=AB+K(A-B)+K^2\ne (A+KI)(B-KI)$

It is proved that, $A+KI,B-KI$ do not commute with each other.

Hence, option D is false.