Question

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

Hint: Two square matrices$A,B$of any order is said to be commutative if and only if$AB = BA$.
At first, we will consider the matrices of order n and then by trial error method, we will try to find out which option will be true.

Complete step by step answer:
It is given in the question that, $A,B$ are two square matrices of order$n$.
Also given that,$A,B$commutes then we have, $AB = BA$
Let us consider the square matrices of order n as,

{{a_{11}}}& \ldots &{{a_{1n}}} \\\ \vdots & \ddots & \vdots \\\ {{a_{n1}}}& \cdots &{{a_{nn}}} \end{array}} \right)$$ and $$B = \left( {\begin{array}{*{20}{c}} {{b_{11}}}& \ldots &{{b_{1n}}} \\\ \vdots & \ddots & \vdots \\\ {{b_{n1}}}& \cdots &{{b_{nn}}} \end{array}} \right)$$ Again it is given that, $$K$$ be any real number. Let us consider$$I$$to be the identity matrix of order$$n$$. Then$$I$$is of the form, $$I = {\left( {\begin{array}{*{20}{c}} 1& \ldots &0 \\\ \vdots & \ddots & \vdots \\\ 0& \cdots &1 \end{array}} \right)_{n \times n}}$$ Now let us try trial and error method and try to find which option will be correct. $$(A - KI)(B - KI)$$ Let us multiply the term we get, $$(A - KI)(B - KI)$$$$ = AB - AKI - BKI + {K^2}I$$ Since it is given that$$A,B$$commutes we will be substituting $$AB = BA$$ in the above equation, Then the equation can be written as, $$ = BA - AKI - BKI + {K^2}I$$ Here now we are going to simplify the above term and write it as the multiplication of two terms, therefore, we get, $$BA - AKI - BKI + {K^2}I$$$$ = (B - KI)(A - KI)$$ Hence finally we have got that $$(A - KI)(B - KI)$$$$ = (B - KI)(A - KI)$$ Which is of the form$$AB = BA$$. Hence, we have found that$$A - KI,B - KI$$ commutes. _Therefore, the correct option is (A) $$A - KI,B - KI$$commute._ Note: Suppose if I is an identity matrix of order$$n$$ and $$K$$ be any real number. Then, $$AKI,BKI$$will also commute if A, B commute each other.