If A and B are square matrices of equal degree, then which one is correct among the followings? (a)

Q: Wait, why isn't matrix multiplication commutative? I thought $A \times B$ should be the same as $B \times A$?

That is a classic trap! Think of matrix multiplication not as simple scaling, but as a row-column dot product. When you calculate $AB$, you are taking rows of $A$ and columns of $B$. When you calculate $BA$, you are taking rows of $B$ and columns of $A$. These are completely different operations! It is like comparing the dot product of vector $u$ and $v$ versus $v$ and $u$—they are the same, but here the structure of the rows and columns changes entirely.

Q: Why can't we just say $A - B = B - A$?

Because subtraction is not commutative! Think about simple numbers: $5 - 3 = 2$, but $3 - 5 = -2$. They are not the same. In matrices, $A - B$ results in elements $a_{ij} - b_{ij}$, while $B - A$ results in $b_{ij} - a_{ij}$. Since $a_{ij} - b_{ij} = -(b_{ij} - a_{ij})$, the matrices are actually negatives of each other, not equal.

Q: Is $A + B = A - B$ ever true?

Only in a very specific case! If you subtract $A$ from both sides, you get $B = -B$, which implies $2B = O$ (the zero matrix). So, this equation only holds if $B$ is the zero matrix. Since the question asks for a property that is universally true for any square matrices, this option is incorrect.

Question

If $A$ and $B$ are square matrices of equal degree, then which one is correct among the followings?

Options:

(A) (a) $A+B=B+A$

(B) (b) $A+B=A-B$

(C) (c) $A-B=B-A$

(D) (d) $AB=BA$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear students. Today, we are not just solving a problem; we are peeling back the layers of one of the most fundamental structures in linear algebra: the matrix.

When you look at a matrix, do not just see a grid of numbers. See a transformation, a data structure, a vessel of information.

We are given two square matrices, $A$ and $B$, of equal degree. Because they share the same dimensions, they are conformable for addition, subtraction, and multiplication.

### The Harmony of Addition

Let us begin with the most intuitive operation: addition. If $A = [a_{ij}]$ and $B = [b_{ij}]$, then the sum $A + B$ is a new matrix where each element is the sum of the corresponding elements: $[a_{ij} + b_{ij}]$.

Think about the nature of these elements $a_{ij}$ and $b_{ij}$. They are real or complex numbers, and the golden rule of scalar addition is commutative.

Since $a_{ij} + b_{ij} = b_{ij} + a_{ij}$ for every position in the matrix, the entire matrix $A + B$ must be equal to $B + A$. This is the **Commutative Law of Addition**, and it is a universal truth in the world of matrices.

**Option (a) stands tall as our correct answer.**

### The Trap of Subtraction

Now, let us address the skeptics. If we define $A - B$, we are looking at the elements $a_{ij} - b_{ij}$.

If we look at $B - A$, we are looking at $b_{ij} - a_{ij}$. As we know from basic arithmetic, $5 - 3 
eq 3 - 5$.

In fact, they are additive inverses, such that $A - B = -(B - A)$. This means subtraction is not commutative; it is a directional operation. This immediately invalidates option (c).

### The Myth of Multiplication

Finally, we arrive at the most complex and often misunderstood operation: matrix multiplication. When we compute $AB$, the element $c_{ij}$ in the product is defined by the summation:

$$c_{ij} = \sum_{k=1}^{n} a_{ik} b_{kj}$$

Now, consider $BA$. Here, we are taking the rows of $B$ and the columns of $A$. The resulting summation is:

$$d_{ij} = \sum_{k=1}^{n} b_{ik} a_{kj}$$

These two summations are fundamentally different. While there are rare, special cases where $AB = BA$, it is not a universal rule. In the general case, **matrix multiplication is strictly non-commutative**. Therefore, option (d) is a trap.

### Conclusion: The Takeaway

We have systematically dismantled the options. We proved that addition is commutative because scalar addition is commutative.

We proved that subtraction is not commutative because of the sign inversion. We proved that multiplication is not commutative because the row-column structure changes entirely when you swap the matrices.

This problem is a beautiful reminder that in mathematics, we must never assume properties carry over from one operation to another. Keep this rigor in your heart, and you will conquer any problem JEE throws your way.

Animated Solution for Mathematics - Matrices and Determinants: If $A$ and $B$ are square matrices of equal degree, then which one is correct among the followings?

Select Answer:

Visualized Solution

Square Matrices $A$ and $B$

Matrix Addition

Commutativity of $A + B$

Matrix Subtraction

Non-Commutativity of $A - B$

Analyzing $A + B = A - B$

Matrix Multiplication

Non-Commutativity of $A B$

Final Conclusion

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Harmony of Addition

The Trap of Subtraction

The Myth of Multiplication

Conclusion

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are square matrices of equal degree, then which one is correct among the followings?

Select Answer:

Visualized Solution

Square Matrices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Matrix Addition

Commutativity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+B

Matrix Subtraction

Non-Commutativity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A−B

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+B=A−B