If A and B are square matrices of size n x n such that A² - B² = (A-B)(A+B), then which of the follo

Q: Wait, why can't I just use the formula $(A-B)(A+B) = A^2 - B^2$ like I do in school algebra?

That is the most common trap! In scalar algebra, multiplication is commutative, meaning $ab = ba$. But in matrix algebra, $AB$ is generally not equal to $BA$. Because the order of multiplication matters, the middle terms $AB$ and $-BA$ do not cancel out unless the matrices commute. Always expand matrix expressions term-by-term to be safe.

Q: Does this mean $A$ and $B$ must be diagonal matrices for the identity to hold?

Not necessarily! While diagonal matrices do commute, they are not the only ones. Any two matrices that satisfy $AB = BA$ will satisfy the given identity. This includes cases where $A$ and $B$ are powers of the same matrix, or where one is the identity matrix, or even specific non-diagonal matrices that happen to commute.

Q: What if $A$ or $B$ is the zero matrix? Does that satisfy the condition?

Yes, it does! If $A$ is the zero matrix, then $0^2 - B^2 = (0-B)(0+B)$, which simplifies to $-B^2 = -B^2$. This is a trivial case where the identity holds because the zero matrix commutes with every matrix. It is a great example of how the general condition $AB = BA$ covers all these special cases.

Question

If $A$ and $B$ are square matrices of size $n 	imes n$ such that $A^2 - B^2 = (A-B)(A+B)$, then which of the following will be always true?

Options:

(A) $A=B$

(B) $AB=BA$

(C) either of $A$ or $B$ is a zero matrix

(D) either of $A$ or $B$ is identity matrix

Sigma Percentile · Accepted Answer

### The Matrix Mirage: Why Intuition Can Be Your Worst Enemy in Algebra

Imagine you are standing at the threshold of a new mathematical world. For years, you have been trained in the elegant, predictable realm of scalar algebra.

You know that $(x-y)(x+y)$ is always $x^2 - y^2$. It is a bedrock truth, a comfort, a constant. But today, we are stepping into the realm of matrices, where that comfort is a dangerous illusion.

In the world of matrices, the order of operations is not just a rule; it is the very soul of the calculation.

### The Trap of Commutativity

Let us look at the problem: we are given $A^2 - B^2 = (A-B)(A+B)$. Your intuition screams that this is true for all $A$ and $B$.

But wait! In matrix algebra, multiplication is not commutative. This means that, in general, $AB 
eq BA$.

This simple fact changes everything. When we multiply two matrices, we are essentially composing linear transformations. Applying transformation $A$ then $B$ is rarely the same as applying $B$ then $A$.

If you rotate a cube and then flip it, you get a different result than if you flip it and then rotate it. This is the physical reality behind the algebraic truth.

### The Anatomy of the Expansion

To solve this, we must strip away our assumptions and perform the expansion manually, with the precision of a surgeon. We start with the right-hand side: $(A-B)(A+B)$.

Using the distributive property, we expand this as:
$$(A-B)(A+B) = A(A+B) - B(A+B)$$

Now, we distribute $A$ and $B$ into the parentheses, being absolutely meticulous about the order:
$$A(A) + A(B) - B(A) - B(B)$$

This gives us:
$$A^2 + AB - BA - B^2$$

Notice what has happened here. We have the $A^2$ and the $-B^2$, just like in scalar algebra, but we are left with the terms $AB - BA$ in the middle. These terms are the ghosts of non-commutativity.

### The Moment of Simplification

Now, let us bring the left-hand side back into the picture. We set our expanded right-hand side equal to the original left-hand side:
$$A^2 - B^2 = A^2 + AB - BA - B^2$$

This is where the magic happens. We can subtract $A^2$ from both sides and add $B^2$ to both sides. The equation collapses, leaving us with:
$$0 = AB - BA$$

This is the core of the problem. For the original identity to hold, the difference between $AB$ and $BA$ must be the zero matrix. In other words, **$AB = BA$**.

### The Final Revelation

We have arrived at our destination. The given identity is not a universal truth for all matrices; it is a specific condition that holds **if and only if the matrices $A$ and $B$ commute**.

This is a profound realization. It teaches us that in advanced mathematics, we cannot rely on the patterns we learned in middle school.

We must derive, we must verify, and we must respect the unique properties of the objects we are studying. Whether you are dealing with matrices, vectors, or tensors, always remember: the order of multiplication is the key to the kingdom.

Animated Solution for Mathematics - Matrices and Determinants: If $A$ and $B$ are square matrices of size $n \times n$ such that $A^{2} - B^{2} = (A - B) (A + B)$ , then which of the following will be always true?

Select Answer:

Visualized Solution

The Given Matrix Equation

Matrix Multiplication Properties

Expanding the RHS

Equating LHS and RHS

Simplifying the Equation

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

The Matrix Mirage

The Trap of Commutativity

The Anatomy of the Expansion

The Moment of Simplification

The Final Revelation

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