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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 5 of 160

Marks: +1, -0

If

A+2 B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}

2 A-B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}

\operatorname{tr}(A) - \operatorname{tr}(B) =

1
2
3
4

Solution:

We have

A+2 B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}

And

2 A-B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}

Now,

2 A+4 B = \begin{bmatrix} 2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2 \end{bmatrix}

Substract Eqs. (ii) from (iii), we get

5 B = \begin{bmatrix} 2 & 4 & 0 \\ 12 & -6 & 6 \\ -10 & 6 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 5 & -5 \\ 10 & -5 & 0 \\ -10 & 5 & 0 \end{bmatrix}

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

\therefore

From Eqs. (ii), we get

\;2 A = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{bmatrix}

\Rightarrow 2 A = \begin{bmatrix} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{bmatrix} \Rightarrow A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix}

\therefore \; \operatorname{tr}(A) - \operatorname{tr}(B)

\; = (1-1+1) - (0-1+0) = 1+1 = 2

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