Let A and B be 3 × 3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. T

Q: Wait, why do we reject the possibility of a unique solution here?

For a homogeneous system of linear equations $PX = O$, a unique solution (the trivial solution $X = O$) exists if and only if the determinant of the coefficient matrix $P$ is non-zero. Since $P$ is a $3 \times 3$ skew-symmetric matrix, its determinant is always zero. Therefore, a unique solution is impossible.

Q: Why is the determinant of an odd-order skew-symmetric matrix always zero?

This is a beautiful property of skew-symmetric matrices. If $P$ is skew-symmetric, then $P^T = -P$. Taking the determinant of both sides, we get $\det(P^T) = \det(-P)$. Since $\det(P^T) = \det(P)$ and $\det(-P) = (-1)^n \det(P)$ (where $n$ is the order of the matrix), we have $\det(P) = (-1)^n \det(P)$. For odd $n$, this becomes $\det(P) = -\det(P)$, which implies $2 \det(P) = 0$, so $\det(P) = 0$.

Q: Can we just assume $A$ and $B$ commute to simplify $P$?

Absolutely not! The problem does not state that $A$ and $B$ commute. Assuming they commute would be a fatal error. We must rely on the general properties of symmetric and skew-symmetric matrices, which hold regardless of whether $A$ and $B$ commute.

Question

Let $A$ and $B$ be $3 	imes 3$ real matrices such that $A$ is symmetric matrix and $B$ is skew-symmetric matrix. Then the system of linear equations $(A^2 B^2 - B^2 A^2) X = O$, where $X$ is a $3 	imes 1$ column matrix of unknown variables and $O$ is a $3 	imes 1$ null matrix, has :

Options:

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler in the realm of linear algebra. Today, we are going to dissect a problem that at first glance might seem like a chaotic mess of matrix products, but beneath the surface lies a beautiful, symmetric structure waiting to be revealed.

We are given two $3 	imes 3$ matrices: $A$, which is symmetric, and $B$, which is skew-symmetric. Our goal is to understand the nature of the solutions to the system $(A^2 B^2 - B^2 A^2) X = O$.

### The Foundation

First, let us ground ourselves in the definitions. We know that $A$ is symmetric, meaning $A^T = A$. We also know $B$ is skew-symmetric, meaning $B^T = -B$.

We define the coefficient matrix $P = A^2 B^2 - B^2 A^2$. The system is $PX = O$. This is a homogeneous system.

The fate of this system—whether it has a unique solution or infinitely many—rests entirely on the determinant of $P$. If $\det(P) 
eq 0$, we have a unique solution. If $\det(P) = 0$, we have infinitely many.

### The Transpose Investigation

To understand $P$, we must look at it from a different angle—literally, by taking its transpose. Let us compute $P^T = (A^2 B^2 - B^2 A^2)^T$.

Using the linearity of the transpose, we distribute it:
$$P^T = (A^2 B^2)^T - (B^2 A^2)^T$$

Now, we invoke the reversal law of transposes: $(MN)^T = N^T M^T$. Applying this, the first term becomes $(B^2)^T (A^2)^T$ and the second becomes $(A^2)^T (B^2)^T$.

So, the expression becomes:
$$P^T = (B^2)^T (A^2)^T - (A^2)^T (B^2)^T$$

### The Revelation

Now, let us simplify the powers. Since $A$ is symmetric, $(A^2)^T = (A^T)^2 = A^2$.

For $B$, since it is skew-symmetric, $(B^2)^T = (B^T)^2 = (-B)^2 = B^2$. Substituting these back into our expression for $P^T$, we get:
$$P^T = B^2 A^2 - A^2 B^2$$

Look closely at this result. It is exactly the negative of our original matrix $P$. Thus, $P^T = -P$. We have just proven that $P$ is a **skew-symmetric matrix**.

### The Final Verdict

Here is the masterstroke. We are working with $3 	imes 3$ matrices, so $P$ is also $3 	imes 3$. The order is $3$, which is odd.

There is a powerful theorem in linear algebra: the determinant of any skew-symmetric matrix of odd order is always zero. Because $\det(P) = 0$, the homogeneous system $PX = O$ cannot have a unique solution.

Instead, it must have infinitely many non-trivial solutions. We have navigated the complexity and arrived at the truth: the system has **infinitely many solutions**. Keep this logic in your toolkit, for it is a powerful weapon in your JEE arsenal.

Select Answer:

Visualized Solution

Identify Matrix Properties

Define the Coefficient Matrix $P$

Apply Transpose to Matrix $P$

Use the Reversal Law of Transpose

Simplify Powers of Transpose

Verify Skew-Symmetry of $P$

Determinant of Odd-Order Skew-Symmetric Matrix

Analyze the Number of Solutions

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Foundation

The Transpose Investigation

The Revelation

The Final Verdict

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The number of matrices in $A$ is

The number of matrices $A$ in $A$ for which the system of linear equations $A x y z = 100$ has a unique solution, is

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Select Answer:

Visualized Solution

Identify Matrix Properties

Define the Coefficient Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Apply Transpose to Matrix .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Use the Reversal Law of Transpose

Simplify Powers of Transpose

Verify Skew-Symmetry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Determinant of Odd-Order Skew-Symmetric Matrix

Analyze the Number of Solutions

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

Analyzing the Setup

The Foundation

The Transpose Investigation

The Revelation

The Final Verdict

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