Question

Let \(A\) and \(B\) be any two \(n\times n\) matrices and \[ tr(A)=\sum_{i=1}^{n}a_{ii}, \qquad tr(B)=\sum_{i=1}^{n}b_{ii}. \] Consider the following statements:
• Statement-I : \(tr(AB)=tr(BA)\).
• Statement-II : \(tr(A+B)=tr(A)+tr(B)\).
• Statement-III : If \(tr(A)=5,\; tr(A^{2})=13\), then \[ tr(A-I)^2=6, \] where \(A_{3\times3}\) and \(I_{3\times3}\) are matrices. Choose the correct option.

• A. Statement-I only is true
• B. Statement-I and Statement-II only are true
• C. Statement-I and Statement-III only are true
• D. Statement-I, Statement-II and Statement-III are all true

Hint:

Remember these four trace identities: \[ tr(A+B)=tr(A)+tr(B) \] \[ tr(kA)=k\,tr(A) \] \[ tr(AB)=tr(BA) \] \[ tr(I_n)=n \] Also, \[ tr\!\left((A-I)^2\right) = tr(A^2)-2tr(A)+n, \] where \(n\) is the order of the identity matrix. These formulas are frequently asked in GATE, CUET(PG), NET and Engineering Mathematics examinations.

Answer 1

The Correct Option is D

Concept: The trace of a square matrix is the sum of its principal diagonal elements. For an \(n\times n\) matrix, \[ tr(A)=\sum_{i=1}^{n}a_{ii}. \] Some important properties of trace are: \[ \boxed{tr(A+B)=tr(A)+tr(B)} \] \[ \boxed{tr(kA)=k\,tr(A)} \] \[ \boxed{tr(AB)=tr(BA)} \] provided that both products are defined. Also, \[ (A-I)^2=A^2-2A+I, \] which is frequently used while evaluating the trace of matrix expressions. These standard properties allow us to verify each statement one by one.

Step 1: Verify Statement-I.
Statement-I states that \[ tr(AB)=tr(BA). \] This is a fundamental property of the trace operator. Proof: Suppose \[ AB=C. \] Then, \[ tr(AB) = \sum_{i=1}^{n}(AB)_{ii} = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}b_{ji}. \] Changing the order of summation, \[ = \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ji}a_{ij} = tr(BA). \] Hence, \[ \boxed{tr(AB)=tr(BA)}. \] Therefore, \[ \boxed{\text{Statement-I is True}.} \]

Step 2: Verify Statement-II.
Statement-II states that \[ tr(A+B)=tr(A)+tr(B). \] Using the definition, \[ tr(A+B) = \sum_{i=1}^{n}(a_{ii}+b_{ii}). \] Separating the summation, \[ = \sum_{i=1}^{n}a_{ii} + \sum_{i=1}^{n}b_{ii}. \] Hence, \[ = tr(A)+tr(B). \] Therefore, \[ \boxed{\text{Statement-II is also True}.} \]

Step 3: Verify Statement-III.
Given, \[ tr(A)=5, \] \[ tr(A^2)=13. \] Also, \[ A \text{ is a } 3\times3 \text{ matrix.} \] We have \[ (A-I)^2 = A^2-2A+I. \] Taking trace on both sides, \[ tr(A-I)^2 = tr(A^2)-2tr(A)+tr(I). \] Since \(I\) is a \(3\times3\) identity matrix, \[ tr(I)=3. \] Substitute the given values. \[ = 13-2(5)+3. \] Simplifying, \[ = 13-10+3 = 6. \] Therefore, \[ \boxed{tr(A-I)^2=6.} \] Hence, \[ \boxed{\text{Statement-III is also True}.} \]

Step 4: Draw the final conclusion.
We have proved that \[ \boxed{\text{Statement-I is True}} \] \[ \boxed{\text{Statement-II is True}} \] \[ \boxed{\text{Statement-III is True}} \] Therefore, all three statements are correct. Hence the correct option is \[ \boxed{(D)\text{ Statement-I, Statement-II and Statement-III are all true}.} \]

Step 5: Important examination note.
Questions based on trace generally test whether students remember the standard identities. The following properties are among the most frequently asked: \[ tr(AB)=tr(BA) \] \[ tr(A+B)=tr(A)+tr(B) \] \[ tr(kA)=k\,tr(A) \] However, \[ \boxed{tr(AB)\neq tr(A)\,tr(B)} \] in general. This is a very common misconception in objective examinations.

Step 6: Final Answer.
Since all the three statements are true, \[ \boxed{\text{Correct Option = (4)}} \]