Q. 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 $tr(A) - tr(B)$ has the value equal to
A) 0
B) 1
C) 2
D) None
Explanation
(c)
$A + 2B = \begin{bmatrix}
1 & 2 & 0 \\
6 & - 3 & 3 \\
- 5 & 3 & 1
\end{bmatrix}$ and $2A - B = \ \begin{bmatrix}
2 & - 1 & 5 \\
2 & - 1 & 6 \\
0 & 1 & 2
\end{bmatrix}$
$\Rightarrow tr(A) + 2tr(B) = - 1$
and $2tr(A) - tr(B) = 3$
Let $tr(A) = x$ and tr$(B) = y$. Then,
$x + 2y = - 1\ \$and$\ 2x - y = 3$
Solving, $x = 1\$and$\ y = - 1.$ Hence,
$tr(A) - tr(B) = x - y = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.