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 trace property.
We know that \[ \mathrm{Tr}(A^2-A)=\mathrm{Tr}(A^2)-\mathrm{Tr}(A) \] So first we calculate \(A^2\).
Given \[ A= \begin{bmatrix} 1 & 1 & 3\\ 1 & 7 & 9\\ 2 & 3 & 7 \end{bmatrix} \]

Step 2: Compute \(A^2=A\cdot A\).
Multiplying the matrices, \[ A^2= \begin{bmatrix} 1 & 1 & 3\\ 1 & 7 & 9\\ 2 & 3 & 7 \end{bmatrix} \begin{bmatrix} 1 & 1 & 3\\ 1 & 7 & 9\\ 2 & 3 & 7 \end{bmatrix} \] Now calculate the diagonal elements because only they are needed for trace.
First diagonal element: \[ (1)(1)+(1)(1)+(3)(2)=1+1+6=8 \] Second diagonal element: \[ (1)(1)+(7)(7)+(9)(3)=1+49+27=77 \] Third diagonal element: \[ (2)(3)+(3)(9)+(7)(7)=6+27+49=82 \] Therefore, \[ \mathrm{Tr}(A^2)=8+77+82 \] So, \[ \mathrm{Tr}(A^2)=167 \]

Step 3: Find \(\mathrm{Tr}(A)\).
The trace of a matrix is the sum of its diagonal elements.
Hence, \[ \mathrm{Tr}(A)=1+7+7 \] Thus, \[ \mathrm{Tr}(A)=15 \]

Step 4: Calculate \(\mathrm{Tr}(A^2-A)\).
Using \[ \mathrm{Tr}(A^2-A)=\mathrm{Tr}(A^2)-\mathrm{Tr}(A), \] we get \[ \mathrm{Tr}(A^2-A)=167-15 \] Therefore, \[ \mathrm{Tr}(A^2-A)=152 \]

Step 5: Final conclusion.
Hence, \[ \boxed{152} \]