Question

If \[ A= \begin{bmatrix} 1 & 1 & 3\\ 1 & 7 & 9\\ 2 & 3 & 7 \end{bmatrix} \] then \[ \mathrm{Tr}(A^2-A)= \] is equal to:

• A. \(0\)
• B. \(-12\)
• C. \(152\)
• D. \(125\)

Hint:

For any square matrices, \[ \mathrm{Tr}(A+B)=\mathrm{Tr}(A)+\mathrm{Tr}(B) \] and \[ \mathrm{Tr}(A-B)=\mathrm{Tr}(A)-\mathrm{Tr}(B) \] Also, the trace of a matrix is always the sum of its principal diagonal elements.

Answer 1

The Correct Option is C

Step 1: Use the linearity of the trace.
The trace $ \mathrm{Tr} $ is linear, so $ \mathrm{Tr}(A^2 - A) = \mathrm{Tr}(A^2) - \mathrm{Tr}(A) $. We only need the diagonal entries of $ A^2 $.
Step 2: Identify the diagonal entries of A.
$ A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix} $. The diagonal is $ 1, 7, 7 $. So $ \mathrm{Tr}(A) = 1 + 7 + 7 = 15 $.
Step 3: Compute the (1,1) entry of A squared.
Row 1 of $ A $ is $ [1, 1, 3] $; Column 1 of $ A $ is $ [1, 1, 2]^T $. Dot product: $ 1(1) + 1(1) + 3(2) = 1 + 1 + 6 = 8 $.
Step 4: Compute the (2,2) entry of A squared.
Row 2 of $ A $ is $ [1, 7, 9] $; Column 2 of $ A $ is $ [1, 7, 3]^T $. Dot product: $ 1(1) + 7(7) + 9(3) = 1 + 49 + 27 = 77 $.
Step 5: Compute the (3,3) entry of A squared.
Row 3 of $ A $ is $ [2, 3, 7] $; Column 3 of $ A $ is $ [3, 9, 7]^T $. Dot product: $ 2(3) + 3(9) + 7(7) = 6 + 27 + 49 = 82 $.
Step 6: Find the final answer.
$ \mathrm{Tr}(A^2) = 8 + 77 + 82 = 167 $. Therefore: \[ \mathrm{Tr}(A^2 - A) = 167 - 15 = 152 \] \[ \boxed{152} \]