Question

If A and B are square matrices of the same order, explain, why in general? ${\left( {A - B} \right)^2} \ne {A^2} - 2AB + {B^2}$

Answer 1

Hint: We can take two arbitrary square matrices A and B. then we can expand the LHS using matrix multiplication. Then we can prove that the LHS is not equal to the RHS as the matrix multiplication is not commutative.

Complete step-by-step answer:
Let A and B be square matrices of any order. We need to prove that ${\left( {A - B} \right)^2} \ne {A^2} - 2AB + {B^2}$
Now we can take the LHS.
$ \Rightarrow LHS = {\left( {A - B} \right)^2}$
We know that ${a^2} = a \times a$
$ \Rightarrow LHS = \left( {A - B} \right)\left( {A - B} \right)$
On expanding the bracket, we get
$ \Rightarrow LHS = A\left( {A - B} \right) - B\left( {A - B} \right)$
On expanding we get,
$ \Rightarrow LHS = {A^2} - AB - BA + {B^2}$
We know that matrix multiplication is not commutative,
$ \Rightarrow AB \ne BA$
Hence, we have
$ \Rightarrow {A^2} - AB - BA + {B^2} \ne {A^2} - 2AB + {B^2}$
So we can say that, $LHS \ne RHS$
$ \Rightarrow {\left( {A - B} \right)^2} \ne {A^2} - 2AB + {B^2}$
So for any square matrix A and B of the same order, ${\left( {A - B} \right)^2} \ne {A^2} - 2AB + {B^2}$.

Note: The identity we disproved here is a well-known identity of algebra. This identity is not applicable to matrices because matrix multiplication is not commutative. If an operation is commutative, it gives the same value if the order of the quantities are changed. We cannot do this question by taking an example because we are asked to prove it generally of square matrices of any order. If we take an example, we can prove the result only for that particular order.