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3 votes
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If $M$ is a $3 \times 3$ matrix such that $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M=\begin{bmatrix}1 & 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}3 & 4 & 5 \end{bmatrix} M = \begin{bmatrix}0 & 1 & 0 \end{bmatrix}$ then $\begin{bmatrix}6 & 7 & 8 \end{bmatrix}M$ is equal to

  1. $\begin{bmatrix}2 & 1 & -2 \end{bmatrix}$
  2. $\begin{bmatrix}0 & 0 & 1 \end{bmatrix}$
  3. $\begin{bmatrix} -1 & 2 & 0 \end{bmatrix}$
  4. $\begin{bmatrix} 9 & 10 & 8 \end{bmatrix}$
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3 Answers

7 votes
7 votes

Here it's given that $\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M=\begin{bmatrix}1 & 0 & 0 \end{bmatrix} \tag{i}$

and

$\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}M=\begin{bmatrix}0 & 1 & 0 \end{bmatrix} \tag{ii}$

 

Now $2\times\mathrm{no(ii)-\mathrm{no(i)}}\Rightarrow$

$\begin{align}  2\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}M-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix}M &=2\begin{bmatrix}0 & 1 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \left( 2\begin{bmatrix} 3 & 4 & 5 \end{bmatrix}-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \right)M &=2\begin{bmatrix}0 & 1 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \left( \begin{bmatrix} 6 & 8 & 10 \end{bmatrix}-\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} \right)M &=\begin{bmatrix}0 & 2 & 0 \end{bmatrix}-\begin{bmatrix}1 & 0 & 0 \end{bmatrix}\\\Rightarrow \begin{bmatrix} (6-0) & (8-1) & (10-2) \end{bmatrix}M &=\begin{bmatrix}(0-1) & (2-0) & (0-0) \end{bmatrix}\\\therefore \begin{bmatrix} 6 & 7 & 8 \end{bmatrix}M &=\begin{bmatrix}-1 & 2 & 0 \end{bmatrix}\end{align}$

 

 

So the correct answer is C.

0 votes
0 votes

 

  c is correct answer .

HINT: just assume M as ' a to i' row wise , setup the equations . you can solve last column of answer easily and the only option that matches with it is c only i.e 0

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