If \[A = \left[ {\begin{array}{*{20}{c}}
  1 \\
  3
\end{array}\begin{array}{*{20}{c}}
  2 \\
  { - 1}
\end{array}\begin{array}{*{20}{c}}
  x \\
  2
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
  y \\
  x \\
  1
\end{array}} \right]\]be such that \[AB = \left[ {\begin{array}{*{20}{c}}
  6 \\
  8
\end{array}} \right]\]then:
A. \[y = 2x\]
B. \[y = - 2x\]
C. \[y = x\]
D. \[y = - x\]

Hint: We use the concept of matrix multiplication and multiply A and B and then equate the elements to the given elements of AB matrix. Find both the values of x and y and relate them for the answer.
* Two matrices can be multiplied if and only if the number of columns of the first matrix is equal to the number of rows of the second matrix.

Complete step-by-step answer:
Here \[A = \left[ {\begin{array}{*{20}{c}}
  1 \\
  3
\end{array}\begin{array}{*{20}{c}}
  2 \\
  { - 1}
\end{array}\begin{array}{*{20}{c}}
  x \\
  2
\end{array}} \right]\] is a matrix of order \[2 \times 3\] which means it has three columns
and \[B = \left[ {\begin{array}{*{20}{c}}
  y \\
  x \\
  1
\end{array}} \right]\]is a matrix of order \[3 \times 1\]which means it has three number of rows.
Since number of columns of A is equal to number of rows in B, we can apply matrix multiplication to AB
Now we calculate the product of two matrices A and B
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  1 \\
  3
\end{array}\begin{array}{*{20}{c}}
  2 \\
  { - 1}
\end{array}\begin{array}{*{20}{c}}
  x \\
  2
\end{array}} \right] \times \left[ {\begin{array}{*{20}{c}}
  y \\
  x \\
  1
\end{array}} \right]\]
From matrix multiplication we know that the first entry of the product matrix will be the sum of product of corresponding elements of first row with the corresponding elements of first the column, we proceed similarly for the second row.
\[ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {1 \times y + 2 \times x + x \times 1} \\
  {3 \times y - 1 \times x + 2 \times 1}
\end{array}} \right]\]
Multiplying the elements
\[
   \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {y + 2x + x} \\
  {3y - x + 2}
\end{array}} \right] \\
   \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {y + 3x} \\
  {3y - x + 2}
\end{array}} \right] \\
 \]
Now we are given in the question that \[AB = \left[ {\begin{array}{*{20}{c}}
  6 \\
  8
\end{array}} \right]\]
Therefore, \[ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {y + 3x} \\
  {3y - x + 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  6 \\
  8
\end{array}} \right]\]
Two matrices are said to be equal if their corresponding elements are equal.
For the first element
\[y + 3x = 6\]
Shifting the value of \[3x\] to RHS of the equation we get
\[y = 6 - 3x\] … (1)
Now for the second term
\[3y - x + 2 = 8\]
Shift 2 to RHS of the equation
\[
  3y - x = 8 - 2 \\
  3y - x = 6 \\
 \]
Now substitute the value of y from equation (1)
\[
   \Rightarrow 3(6 - 3x) - x = 6 \\
   \Rightarrow 18 - 9x - x = 6 \\
   \Rightarrow - 10x + 18 = 6 \\
 \]
Shifting the constant to one side of the equation
\[
   \Rightarrow - 10x = 6 - 18 \\
   \Rightarrow - 10x = - 12 \\
 \]
Cancel out the negative sign from both sides of the equation
\[ \Rightarrow 10x = 12\]
Divide both sides by 10
\[ \Rightarrow \dfrac{{10x}}{{10}} = \dfrac{{12}}{{10}}\]
Cancel same terms from numerator and denominator
\[ \Rightarrow x = \dfrac{{12}}{{10}}\]
Now we find the value of y using equation (1)
\[
   \Rightarrow y = 6 - 3(\dfrac{{12}}{{10}}) \\
   \Rightarrow y = 6 - \dfrac{{36}}{{10}} \\
 \]
Take LCM on RHS of the equation
\[
   \Rightarrow y = \dfrac{{60 - 36}}{{10}} \\
   \Rightarrow y = \dfrac{{24}}{{10}} \\
 \]
So, \[x = \dfrac{{12}}{{10}}\]and \[y = \dfrac{{24}}{{10}}\]
We can write \[y = \dfrac{{24}}{{10}} = 2(\dfrac{{12}}{{10}}) = 2x\]
So, \[y = 2x\]

So, option A is correct.

Note: Students are likely to make mistakes while shifting the values from one side of the equation to the other side of the equation as they forget to change sign while shifting, always keep in mind the sign changes when we shift a value from one side of the equation to another.