Question:

If $$ A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} $$ then $ AA^T $ is:

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The product \( AA^T \) of any matrix \( A \) with its transpose is always a symmetric matrix.
Updated On: May 21, 2025
  • Symmetric matrix
  • Skew-Symmetric matrix
  • Singular matrix
  • Inverse of A
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The Correct Option is A

Solution and Explanation

Let’s compute \( AA^T \). First, compute \( A^T \) (transpose of \( A \)): \[ A^T = \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix} \] Now, compute the product \( AA^T \): \[ AA^T = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix} \] Perform matrix multiplication: \[ AA^T = \begin{bmatrix} % Option (3)(3) + (-3)(-3) + (4)(4) & (3)(2) + (-3)(-3) + (4)(4) & (3)(0) + (-3)(-1) + (4)(1) \\ % Option (2)(3) + (-3)(-3) + (4)(4) & (2)(2) + (-3)(-3) + (4)(4) & (2)(0) + (-3)(-1) + (4)(1) \\ % Option (0)(3) + (-1)(-3) + (1)(4) & (0)(2) + (-1)(-3) + (1)(4) & (0)(0) + (-1)(-1) + (1)(1) \end{bmatrix} \] Calculate values: \[ AA^T = \begin{bmatrix} 9 + 9 + 16 & 6 + 9 + 16 & 0 + 3 + 4 \\ 6 + 9 + 16 & 4 + 9 + 16 & 0 + 3 + 4 \\ 0 + 3 + 4 & 0 + 3 + 4 & 0 + 1 + 1 \end{bmatrix} = \begin{bmatrix} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{bmatrix} \] Observe that: \[ AA^T = (AA^T)^T \] So the matrix is symmetric.
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