Question

If \( A = \begin{bmatrix} \alpha & 3 & 1 2 & \beta & 6 -2 & -1 & \gamma \end{bmatrix} \) and \( AA^T = \begin{bmatrix} 35 & 28 & -13 28 & 56 & -8 -13 & -8 & 5 \end{bmatrix} \), then Trace of \( A = \)

• A. 12
• B. 14
• C. 9
• D. 11

Hint:

In \( AA^T \), the diagonal entry \( d_i \) is always the square of the norm of the \( i \)-th row vector. Use this to find absolute values of unknowns quickly, then use one off-diagonal entry to fix the signs.

Answer 1

The Correct Option is C

Concept: The product of a matrix \( A \) and its transpose \( A^T \) results in a symmetric matrix where the diagonal elements are the sums of the squares of the elements in the corresponding rows of \( A \).
• \( (AA^T)_{ii} = \sum (\text{elements of row } i)^2 \)
• Trace of \( A = \text{Sum of diagonal elements} = \alpha + \beta + \gamma \).

Step 1: Equating the diagonal elements of \( AA^T \).
From the first diagonal element: \[ \alpha^2 + 3^2 + 1^2 = 35 \Rightarrow \alpha^2 + 10 = 35 \Rightarrow \alpha^2 = 25 \Rightarrow \alpha = \pm 5 \] From the second diagonal element: \[ 2^2 + \beta^2 + 6^2 = 56 \Rightarrow \beta^2 + 40 = 56 \Rightarrow \beta^2 = 16 \Rightarrow \beta = \pm 4 \] From the third diagonal element: \[ (-2)^2 + (-1)^2 + \gamma^2 = 5 \Rightarrow 5 + \gamma^2 = 5 \Rightarrow \gamma^2 = 0 \Rightarrow \gamma = 0 \]

Step 2: Determining signs using off-diagonal elements.
Consider the element at \( (1, 2) \) of \( AA^T \): \[ \alpha(2) + 3(\beta) + 1(6) = 28 \Rightarrow 2\alpha + 3\beta = 22 \] If \( \alpha = 5 \) and \( \beta = 4 \): \( 2(5) + 3(4) = 10 + 12 = 22 \). (This satisfies the condition). Checking \( (1, 3) \): \( \alpha(-2) + 3(-1) + 1(0) = -10 - 3 = -13 \). (Matches).

Step 3: Calculating the Trace.
\[ \text{Trace}(A) = \alpha + \beta + \gamma = 5 + 4 + 0 = 9 \]