If A = begin{bmatrix} 1 & 2 & x 3 & -1 & 2 e | vedclass

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(A) Given $A = \begin{bmatrix} 1 & 2 & x \ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \ x \ 1 \end{bmatrix}$.Calculating the product $AB$

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$y = 2x$

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(A) Given $A = \begin{bmatrix} 1 & 2 & x \ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \ x \ 1 \end{bmatrix}$.Calculating the product $AB$:$AB = \begin{bmatrix} 1 & 2 & x \ 3 & -1 & 2 \end{bmatrix} \begin{bmatrix} y \ x \ 1 \end{bmatrix} = \begin{bmatrix} 1(y) + 2(x) + x(1) \ 3(y) - 1(x) + 2(1) \end{bmatrix} = \begin{bmatrix} y + 3x \ 3y - x + 2 \end{bmatrix}$.Equating this to the given matrix $\begin{bmatrix} 6 \ 8 \end{bmatrix}$:$y + 3x = 6$  (Equation $1$)$3y - x + 2 = 8 \Rightarrow 3y - x = 6$  (Equation $2$)From Equation $1$,$y = 6 - 3x$. Substituting this into Equation $2$:$3(6 - 3x) - x = 6$$18 - 9x - x = 6$$18 - 10x = 6$$10x = 12 \Rightarrow x = \frac{12}{10} = \frac{6}{5}$.Substituting $x = \frac{6}{5}$ into $y = 6 - 3x$:$y = 6 - 3(\frac{6}{5}) = 6 - \frac{18}{5} = \frac{30 - 18}{5} = \frac{12}{5}$.Comparing $x$ and $y$,we see that $y = 2x$ (since $\frac{12}{5} = 2 	imes \frac{6}{5}$).Thus,the correct option is $A$.

If $A = \begin{bmatrix} 1 & 2 & x \\ 3 & -1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} y \\ x \\ 1 \end{bmatrix}$ are such that $AB = \begin{bmatrix} 6 \\ 8 \end{bmatrix}$,then:

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