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NCERT Exemplar · Q70

Q.If AA is a symmetric matrix, then A3A^3 is a _________ matrix.

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A symmetric matrix satisfies AT=AA^T = A. When you cube it, the transpose becomes (A3)T=(AT)3=A3(A^3)^T = (A^T)^3 = A^3, so A3A^3 is also symmetric. The answer is symmetric.

The key here is to understand what "symmetric" means in matrix terms, and then see how that property behaves under multiplication.

A matrix AA is called symmetric if it equals its own transpose: AT=AA^T = A. Geometrically, this means the matrix is "mirrored" across its main diagonal — the (i,j)(i,j) entry equals the (j,i)(j,i) entry for all i,ji, j.

Now, the question asks about A3A^3. The natural instinct is to check: does the symmetry survive when we multiply the matrix by itself repeatedly? The answer is yes, and the reason lies in a simple property of transposes.

For any two matrices AA and BB (of compatible sizes), (AB)T=BTAT(AB)^T = B^T A^T.

This is the "reverse order" rule for transposes. It's the only tool we need.

Let's work through it step by step.

  1. Start with what we know.

    We are given that AA is symmetric, so AT=AA^T = A.

  2. Consider A3A^3.

    By definition, A3=A⋅A⋅AA^3 = A \cdot A \cdot A (matrix multiplication).

  3. Take the transpose of A3A^3.

    Using the reverse-order rule repeatedly:

(A3)T=(A⋅A⋅A)T=AT⋅AT⋅AT(A^3)^T = (A \cdot A \cdot A)^T = A^T \cdot A^T \cdot A^T

The order reverses: the last AA becomes first, and so on.

  1. Substitute the symmetry condition. Since AT=AA^T = A, we replace each ATA^T with AA:

(A3)T=A⋅A⋅A=A3(A^3)^T = A \cdot A \cdot A = A^3

  1. Interpret the result. …

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