Key Concept: Non-Commutativity of Matrix Multiplication

For real numbers $x$ and $y$, the identity $(x-y)(x+y) = x^2 - y^2$ holds true. However, when dealing with matrices, this identity does not generally hold. The fundamental reason is that matrix multiplication is not commutative in general, meaning for two matrices $A$ and $B$, $AB$ is usually not equal to $BA$. The order of multiplication matters significantly.

The identity $(A-B)(A+B) = A^2 - B^2$ holds for matrices $A$ and $B$ if and only if $A$ and $B$ commute, i.e., $AB = BA$. This problem essentially asks us to find the condition under which the given equation is true.

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Step-by-Step Derivation

We are given that $A$ and $B$ are square matrices of size $n \times n$ and satisfy the equation: ${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$

Our goal is to analyze this equation and determine which of the given options must always be true.

1. Expand the Right-Hand Side (RHS): We need to carefully expand the product $\left( {A - B} \right)\left( {A + B} \right)$ by applying the distributive property of matrix multiplication, while strictly maintaining the order of matrices. $\left( {A - B} \right)\left( {A + B} \right) = A(A+B) - B(A+B)$ This is the first step of distribution. We distribute $(A-B)$ across $(A+B)$.

2. Further Distribute: Now, distribute $A$ and $-B$ into their respective parentheses: $= A \cdot A + A \cdot B - B \cdot A - B \cdot B$ It is crucial here to note that $A \cdot B$ and $B \cdot A$ are kept distinct. We do not assume they are equal.

3. Simplify the terms: $= A^2 + AB - BA - B^2$ This is the expanded form of the RHS.

4. Substitute the expanded RHS back into the given equation: Now, equate the given LHS ($A^2 - B^2$) with our expanded RHS: ${A^2} - {B^2} = A^2 + AB - BA - B^2$

5. Simplify the equation: We can subtract $A^2$ from both sides of the equation: $- {B^2} = AB - BA - B^2$ Next, we can add $B^2$ to both sides of the equation: $0 = AB - BA$

6. Conclude the condition: From $0 = AB - BA$, we can rearrange the terms to get: $AB = BA$

   This means that the given condition, ${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$, is true if and only if matrices $A$ and $B$ commute, i.e., $AB = BA$. Therefore, if the given equation holds, then $AB=BA$ must always be true.

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Analysis of Options

We have established that the given condition implies $AB=BA$. Now let's check the options:

• (A) $A=B$: If $A=B$, then $AB = A \cdot A = A^2$ and $BA = A \cdot A = A^2$, so $AB=BA$ is true. In this case, the given equation becomes $A^2 - A^2 = (A-A)(A+A)$, which simplifies to $0 = 0$, which is always true. So $A=B$ is a sufficient condition for the given equation to hold. However, it is not a necessary condition. For example, consider $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix}$. Here $AB=BA$ (both equal $\begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix}$), and the given equation holds, but $A \neq B$. Thus, $A=B$ is not always true.

• (B) $AB=BA$: As derived in our step-by-step process, the given equation ${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$ simplifies directly to $AB=BA$. This means that whenever the given equation is true, the condition $AB=BA$ must also be true. Therefore, this option is always true.

• (C) either of $A$ or $B$ is a zero matrix: If $A=0$, then the equation becomes $-B^2 = (-B)(B) = -B^2$, which is true. If $B=0$, then the equation becomes $A^2 = A \cdot A = A^2$, which is true. So, if either $A$ or $B$ is a zero matrix, the given condition holds. In both these cases, $AB=BA$ also holds (e.g., $0 \cdot B = 0$ and $B \cdot 0 = 0$). However, $A$ and $B$ don't have to be zero matrices for $AB=BA$ to hold, or for the original equation to hold (e.g., diagonal matrices, as shown above). So this is not always true.

• (D) either of $A$ or $B$ is identity matrix: Similar to the zero matrix case, if $A=I$ (identity matrix), then $I^2-B^2 = (I-B)(I+B) \implies I-B^2 = I+B-B-B^2 \implies I-B^2 = I-B^2$, which is true. Also, $IB=BI=B$, so $AB=BA$ holds. Similarly for $B=I$. However, this is not a general requirement for the given equation to hold. So this is not always true.

Therefore, the only statement that is always true when the given equation holds is $AB=BA$.

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Common Mistakes and Tips

• Assuming Commutativity: The most common mistake in matrix algebra problems is to assume that $AB=BA$. Always remember that matrix multiplication is generally not commutative.
• Order of Multiplication: When expanding matrix expressions like $(A-B)(A+B)$, meticulously maintain the order of multiplication. $AB$ is different from $BA$.
• Scalar vs. Matrix Algebra: Be aware of the differences between algebraic identities for scalars and matrices. Identities involving products often behave differently in matrix algebra due to non-commutativity.

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Summary and Key Takeaway

The given equation ${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$ is a specific condition that holds true for matrices $A$ and $B$ if and only if they commute. By expanding the right-hand side $(A-B)(A+B)$ as $A^2 + AB - BA - B^2$ and equating it to $A^2 - B^2$, we directly derive the condition $AB - BA = 0$, which means $AB = BA$. Thus, if the given equality holds, it is always true that $A$ and $B$ commute.

The final answer is $\boxed{\text{AB=BA}}$.