Question

If A and B are square matrices of order \[n \times n\], then \[{(A - B)^2}\]is equal
A. \[{A^2} - {B^2}\]
B. \[{A^2} - 2AB + {B^2}\]
C. \[{A^2} + 2AB + {B^2}\]
D. \[{A^2} - AB - BA + {B^2}\]

Answer 1

Hint: In the given problem we consider two arbitrary matrices A and B of order \[n \times n\]. We expand \[{(A - B)^2}\] using matrix multiplication and hence obtain the required result. Since matrix multiplication is not commutative,\[AB \ne BA\].

Complete step by step Solution:
We are given that A and B are square matrices of order \[n \times n\]
We know that \[{(A - B)^2} = (A - B)(A - B)\] since \[{a^2} = a \times a\]
On expanding we get,
\[{(A - B)^2} = A(A - B) - B(A - B)\]
\[ \Rightarrow {(A - B)^2} = {A^2} - AB - BA + {B^2}\] (Since matrix multiplication is not commutative,\[AB \ne BA\])
Hence, If A and B are square matrices of order \[n \times n\], then \[{(A - B)^2} = {A^2} - AB - BA + {B^2}\]

Therefore, the correct option is (D).

Note:To solve the given problem, one must know the basic properties of matrices. The matrix system does not necessarily follow all rules of the algebra. If an operation is commutative, it gives the same value if the order of the quantities is changed. It is very important to note that Matrix multiplication is not commutative i.e., \[AB \ne BA\]. Therefore \[{(A - B)^2} \ne {A^2} - 2AB + {B^2}\].