Determinants Matrices Question 133
Question: Let $ A+2B= \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \\ \end{bmatrix} $ and $ 2A-B= \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \\ \end{bmatrix} $ , then $\operatorname{tr}(A) - \operatorname{tr}(B)$ is
Options:
A) 1
B) 3
C) 2
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Here to find the value of $ tr(A)-tr(B), $ we need not to find the matrices A and B. We can find tr(A)-tr[b] using the properties of trace of matrix, i.e., $ A+2B= \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \\ \end{bmatrix} \Rightarrow tr(A+2B)=-1(1) $ Or $ tr(A)+2tr(B)=-1 $
$ \Rightarrow 2A-B= \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \\ \end{bmatrix} $
$ \Rightarrow tr(2A-B)=3 $ or $ 2tr(A)-tr(B)=3(2) $ Solving (1) and (2), we get tr[a] = 1 and tr[b] = -1
$ \Rightarrow tr(A)-tr(B)=2 $
BETA
