Question:
Let 𝑀 be a 3×3 real matrix. If 𝑃=𝑀+𝑀𝑇 and 𝑄=𝑀-𝑀𝑇, then which of the following statements is/are always TRUE?
Let 𝑀 be a 3×3 real matrix. If 𝑃=𝑀+𝑀𝑇 and 𝑄=𝑀-𝑀𝑇, then which of the following statements is/are always TRUE?
Updated On: Nov 17, 2025
- det(𝑃 2𝑄 3 )=0
- trace(𝑄+𝑄 2 )=0
- 𝑋 𝑇𝑄 2𝑋=0, for all 𝑋∈ℝ3
- 𝑋 𝑇𝑃𝑋 = 2𝑋 𝑇𝑀𝑋, for all 𝑋 ∈ ℝ3
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The Correct Option is A, D
Solution and Explanation
To solve this question, we will analyze each given statement one by one to determine its validity with the provided conditions.
Let's first understand the terms:
- \(P = M + M^T\): This is a symmetric matrix because the sum of a matrix and its transpose is always symmetric.
- \(Q = M - M^T\): This matrix is skew-symmetric because the difference of a matrix and its transpose is always skew-symmetric. For any skew-symmetric matrix \(A\), the property \((A^T = -A)\) holds, and also for odd-order skew-symmetric matrices, \(\det(A)=0\).
Now, let's evaluate each statement:
- Statement: \(\det(P^2Q^3) = 0\)
- We know that \(\det(Q) = 0\) because \(Q\) is a 3×3 skew-symmetric matrix.
- Since \(Q^3\) is also a skew-symmetric, odd-order matrix, it maintains the property \(\det(Q^3) = 0\).
- Therefore, \(\det(P^2Q^3) = \det(P^2) \cdot \det(Q^3) = \det(P^2) \cdot 0 = 0\).
- This statement is true.
- Statement: \(\text{trace}(Q + Q^2) = 0\)
- The trace of any skew-symmetric matrix \(Q\) is zero because \(Q + Q^T = 0\) implies \(\text{trace}(Q) = -\text{trace}(Q)\).
- However, the trace of \(Q^2\) is not necessarily zero. Hence \(\text{trace}(Q^2) \neq 0\) in general.
- Thus, this statement is not always true.
- Statement: \(X^TQ^2X = 0, \text{ for all } X \in \mathbb{R}^3\)
- For any skew-symmetric matrix \(Q\), \(Q^2\) is symmetric. However, there is no guarantee that the quadratic form \(X^TQ^2X\) will always result in zero.
- This is not true for all \(X\), so the statement is false.
- Statement: \(X^T P X = 2X^T M X, \text{ for all } X \in \mathbb{R}^3\)
- Calculate \(X^T P X = X^T (M + M^T) X = X^T M X + X^T M^T X\).
- Since \(X^T M^T X = (X^T M X)^T = X^T M X\), we have \(X^T P X = 2X^T M X\).
- This statement is true.
Based on the analysis, the true statements are:
- \(\det(P^2Q^3) = 0\)
- \(X^T P X = 2 X^T M X, \text{ for all } X \in \mathbb{R}^3\)
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