Question:
Let \(A\) be a \(3\times3\) matrix such that
\[
AA^{T}=I_3.
\]
Then \(A\) is:
Let \(A\) be a \(3\times3\) matrix such that
\[
AA^{T}=I_3.
\]
Then \(A\) is:
Show Hint
Whenever you see \(AA^T=I\) or \(A^TA=I\), immediately identify the matrix as orthogonal. This is one of the most frequently tested matrix identities.
Updated On: Jun 10, 2026
- Singular matrix
- Orthogonal matrix
- Skew-symmetric matrix
- Nilpotent matrix
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The Correct Option is B
Solution and Explanation
Concept:
An orthogonal matrix is a square matrix whose transpose is equal to its inverse.
The defining property is
\[
A^TA=I
\]
or equivalently
\[
AA^T=I.
\]
These conditions imply that the rows and columns form orthonormal vectors.
Step 1: Examine the given condition We are given \[ AA^T=I_3 \] where \(I_3\) denotes the identity matrix of order \(3\).
Step 2: Compare with the definition of orthogonal matrices A matrix \(A\) is orthogonal if \[ A^{-1}=A^T. \] Multiplying both sides by \(A\), \[ AA^T=I. \] This is exactly the condition given in the question.
Step 3: Verify the conclusion Since \[ AA^T=I_3, \] it follows immediately that \[ A^{-1}=A^T. \] Therefore \(A\) satisfies the definition of an orthogonal matrix. Hence, \[ \boxed{A \text{ is an orthogonal matrix}} \]
Step 4: Eliminate other options A singular matrix has determinant zero. For an orthogonal matrix, \[ |A|=\pm1, \] which is never zero. Therefore \(A\) is not singular. A skew-symmetric matrix satisfies \[ A^T=-A, \] which is not implied here. A nilpotent matrix satisfies \[ A^n=0 \] for some positive integer \(n\), which is impossible for an invertible matrix. Thus the only correct option is (B).
Step 1: Examine the given condition We are given \[ AA^T=I_3 \] where \(I_3\) denotes the identity matrix of order \(3\).
Step 2: Compare with the definition of orthogonal matrices A matrix \(A\) is orthogonal if \[ A^{-1}=A^T. \] Multiplying both sides by \(A\), \[ AA^T=I. \] This is exactly the condition given in the question.
Step 3: Verify the conclusion Since \[ AA^T=I_3, \] it follows immediately that \[ A^{-1}=A^T. \] Therefore \(A\) satisfies the definition of an orthogonal matrix. Hence, \[ \boxed{A \text{ is an orthogonal matrix}} \]
Step 4: Eliminate other options A singular matrix has determinant zero. For an orthogonal matrix, \[ |A|=\pm1, \] which is never zero. Therefore \(A\) is not singular. A skew-symmetric matrix satisfies \[ A^T=-A, \] which is not implied here. A nilpotent matrix satisfies \[ A^n=0 \] for some positive integer \(n\), which is impossible for an invertible matrix. Thus the only correct option is (B).
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