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,
\[A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}& \ldots &{{a_{1n}}} \\
   \vdots & \ddots & \vdots \\
  {{a_{n1}}}& \cdots &{{a_{nn}}}
\end{array}} \right)\] 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.