To solve for matrix [tex]\( X \)[/tex] in the equation [tex]\(\frac{1}{2}(X + Z) = Y\)[/tex], follow these steps:
1. Start with the given equation:
[tex]\[\frac{1}{2}(X + Z) = Y\][/tex]
2. Eliminate the fraction by multiplying both sides of the equation by 2:
[tex]\[X + Z = 2Y\][/tex]
3. Solve for [tex]\( X \)[/tex] by isolating it on one side of the equation:
[tex]\[X = 2Y - Z\][/tex]
4. Substitute the given matrices [tex]\( Y \)[/tex] and [tex]\( Z \)[/tex] into the equation:
[tex]\[Y = \left[\begin{array}{cc}4 & 8 \\ -3 & -6\end{array}\right]\][/tex]
[tex]\[Z = \left[\begin{array}{cc}-1 & -2 \\ 7 & 2\end{array}\right]\][/tex]
5. Compute [tex]\( 2Y \)[/tex] by multiplying each element of [tex]\( Y \)[/tex] by 2:
[tex]\[2Y = 2 \times \left[\begin{array}{cc}4 & 8 \\ -3 & -6\end{array}\right] = \left[\begin{array}{cc}8 & 16 \\ -6 & -12\end{array}\right]\][/tex]
6. Subtract matrix [tex]\( Z \)[/tex] from [tex]\( 2Y \)[/tex] to find [tex]\( X \)[/tex]:
[tex]\[X = \left[\begin{array}{cc}8 & 16 \\ -6 & -12\end{array}\right] - \left[\begin{array}{cc}-1 & -2 \\ 7 & 2\end{array}\right] = \left[\begin{array}{cc}8+1 & 16+2 \\ -6-7 & -12-2\end{array}\right] = \left[\begin{array}{cc}9 & 18 \\ -13 & -14\end{array}\right]\][/tex]
Therefore, matrix [tex]\( X \)[/tex] is:
[tex]\[
\left[\begin{array}{cc}9 & 18 \\ -13 & -14\end{array}\right]
\][/tex]
So the correct option is:
[tex]\[
\left[\begin{array}{cc}9 & 18 \\ -13 & -14\end{array}\right]
\][/tex]
