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Question: Let A + 2B =\(egin{bmatrix} 1 & 2 & 0 \\ 6 & - 3 & 3 \\ - 5 & 3 & 1 \end{bmatrix}\)and 2A –B = \(...

Let A + 2B =$egin{bmatrix} 1 & 2 & 0 \ 6 & - 3 & 3 \

  • 5 & 3 & 1 \end{bmatrix}and2A–B=and 2A –B = egin{bmatrix} 2 & - 1 & 5 \ 2 & - 1 & 6 \ 0 & 1 & 2 \end{bmatrix}$then Tr(1) – Tr(2) has the value equal to
A

0

B

1

C

2

D

None of these

Answer

2

Explanation

Solution

2A + 4B = $\begin{bmatrix} 2 & 4 & 0 \ 12 & - 6 & 6 \

  • 10 & 6 & 2 \end{bmatrix}$ … (1)

2A –B = [2−152−16012]\begin{bmatrix} 2 & - 1 & 5 \\ 2 & - 1 & 6 \\ 0 & 1 & 2 \end{bmatrix} … (2)

On solving (1) and (2)

B = $\begin{bmatrix} 0 & 1 & - 1 \ 2 & - 1 & 0 \

  • 2 & 1 & 0 \end{bmatrix}& A =\begin{bmatrix} 1 & 0 & 2 \ 2 & - 1 & 3 \
  • 1 & 1 & 1 \end{bmatrix}$

So Tr (1) – Tr (2) = 1 – (–1) = 2