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Solve the vector equation for a and b: ${\begin{bmatrix}2\\b\\3\end{bmatrix} + \begin{bmatrix}a\\5\\4\end{bmatrix}=\begin{bmatrix}b\\2a\\7\end{bmatrix}}$

I don't understand the question. I am aware of simultaneous equations with the determinant for matrices but this does not seem applicable to this question.

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It just means this -

$2+a=b$ and $b+5=2a$

$a=7$ and $b=9$

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I am sorry if this is an inappropriate question but what was your JEE rank? I'm an aspired. – N.S.JOHN Feb 23 at 13:13
    
@N.S.JOHN You could have asked in chat. Anyway, It's AIR151 in 2012. – Win Vineeth Feb 23 at 13:19
    
Did you go to coaching classes? – N.S.JOHN Feb 23 at 13:21
    
@N.S.JOHN No, I didn't. I solved lot of papers from FITJEE etc.. – Win Vineeth Feb 23 at 13:27

$$\begin{cases} 2+a=b\\ b+5=2a\\ 3+4=7 \end{cases}$$

From the first line: $a=\color{red}{b-2}$ Now $b+5=2\color{red}{(b-2)}$

So $b+5=2b-4\Longrightarrow \boxed{b=9}$ And $\boxed{a=7}$

hope it helps

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You can formulate your equations in matrix form and apply your known solution methods: $$ \begin{bmatrix}2\\b\\3\end{bmatrix} + \begin{bmatrix}a\\5\\4\end{bmatrix}= \begin{bmatrix}b\\2a\\7\end{bmatrix} $$ $$ \left[ \begin{array}{rr} -1 & 1 \\ 2 & -1 \\ \end{array} \right] \begin{bmatrix} a \\ b \\ \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \\ \end{bmatrix} $$ The third equation features no unkown and it is true, so it can be neglected. If it were false the system had no solution.

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