Answer :
To solve the matrix equation for [tex]\( x \)[/tex] and [tex]\( y \)[/tex], follow these steps:
The given matrix equation is:
[tex]\[ \frac{1}{2} \left[\begin{array}{cc} 4 & 8 \\ x+3 & -4 \end{array}\right] - 3 \left[\begin{array}{cc} 1 & y+1 \\ -1 & -2 \end{array}\right] = \left[\begin{array}{cc} -1 & -5 \\ 7 & 4 \end{array}\right] \][/tex]
First, simplify the left side of the equation.
Calculate the scaled matrices:
[tex]\[ \frac{1}{2} \left[\begin{array}{cc} 4 & 8 \\ x+3 & -4 \end{array}\right] = \left[\begin{array}{cc} \frac{4}{2} & \frac{8}{2} \\ \frac{x+3}{2} & \frac{-4}{2} \end{array}\right] = \left[\begin{array}{cc} 2 & 4 \\ \frac{x+3}{2} & -2 \end{array}\right] \][/tex]
[tex]\[ 3 \left[\begin{array}{cc} 1 & y+1 \\ -1 & -2 \end{array}\right] = \left[\begin{array}{cc} 3 \cdot 1 & 3(y+1) \\ 3(-1) & 3(-2) \end{array}\right] = \left[\begin{array}{cc} 3 & 3y+3 \\ -3 & -6 \end{array}\right] \][/tex]
Now, subtract the two matrices:
[tex]\[ \left[\begin{array}{cc} 2 & 4 \\ \frac{x+3}{2} & -2 \end{array}\right] - \left[\begin{array}{cc} 3 & 3y+3 \\ -3 & -6 \end{array}\right] = \left[\begin{array}{cc} 2 - 3 & 4 - (3y + 3) \\ \frac{x+3}{2} + 3 & -2 - (-6) \end{array}\right] = \left[\begin{array}{cc} -1 & 1 - 3y \\ \frac{x+3}{2} + 3 & 4 \end{array}\right] \][/tex]
Set this result equal to the given matrix:
[tex]\[ \left[\begin{array}{cc} -1 & 1 - 3y \\ \frac{x+3}{2} + 3 & 4 \end{array}\right] = \left[\begin{array}{cc} -1 & -5 \\ 7 & 4 \end{array}\right] \][/tex]
Now, match the corresponding elements:
1. Top-left:
[tex]\[ -1 = -1 \][/tex]
This is true, so it is correct.
2. Top-right:
[tex]\[ 1 - 3y = -5 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ 1 - 3y = -5 \\ 1 + 5 = 3y \\ 6 = 3y \\ y = 2 \][/tex]
3. Bottom-left:
[tex]\[ \frac{x+3}{2} + 3 = 7 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{x+3}{2} + 3 = 7 \\ \frac{x+3}{2} = 4 \\ x + 3 = 8 \\ x = 5 \][/tex]
4. Bottom-right:
[tex]\[ 4 = 4 \][/tex]
This is true, so it is correct.
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ \boxed{x = 5, y = 2} \][/tex]
The given matrix equation is:
[tex]\[ \frac{1}{2} \left[\begin{array}{cc} 4 & 8 \\ x+3 & -4 \end{array}\right] - 3 \left[\begin{array}{cc} 1 & y+1 \\ -1 & -2 \end{array}\right] = \left[\begin{array}{cc} -1 & -5 \\ 7 & 4 \end{array}\right] \][/tex]
First, simplify the left side of the equation.
Calculate the scaled matrices:
[tex]\[ \frac{1}{2} \left[\begin{array}{cc} 4 & 8 \\ x+3 & -4 \end{array}\right] = \left[\begin{array}{cc} \frac{4}{2} & \frac{8}{2} \\ \frac{x+3}{2} & \frac{-4}{2} \end{array}\right] = \left[\begin{array}{cc} 2 & 4 \\ \frac{x+3}{2} & -2 \end{array}\right] \][/tex]
[tex]\[ 3 \left[\begin{array}{cc} 1 & y+1 \\ -1 & -2 \end{array}\right] = \left[\begin{array}{cc} 3 \cdot 1 & 3(y+1) \\ 3(-1) & 3(-2) \end{array}\right] = \left[\begin{array}{cc} 3 & 3y+3 \\ -3 & -6 \end{array}\right] \][/tex]
Now, subtract the two matrices:
[tex]\[ \left[\begin{array}{cc} 2 & 4 \\ \frac{x+3}{2} & -2 \end{array}\right] - \left[\begin{array}{cc} 3 & 3y+3 \\ -3 & -6 \end{array}\right] = \left[\begin{array}{cc} 2 - 3 & 4 - (3y + 3) \\ \frac{x+3}{2} + 3 & -2 - (-6) \end{array}\right] = \left[\begin{array}{cc} -1 & 1 - 3y \\ \frac{x+3}{2} + 3 & 4 \end{array}\right] \][/tex]
Set this result equal to the given matrix:
[tex]\[ \left[\begin{array}{cc} -1 & 1 - 3y \\ \frac{x+3}{2} + 3 & 4 \end{array}\right] = \left[\begin{array}{cc} -1 & -5 \\ 7 & 4 \end{array}\right] \][/tex]
Now, match the corresponding elements:
1. Top-left:
[tex]\[ -1 = -1 \][/tex]
This is true, so it is correct.
2. Top-right:
[tex]\[ 1 - 3y = -5 \][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[ 1 - 3y = -5 \\ 1 + 5 = 3y \\ 6 = 3y \\ y = 2 \][/tex]
3. Bottom-left:
[tex]\[ \frac{x+3}{2} + 3 = 7 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ \frac{x+3}{2} + 3 = 7 \\ \frac{x+3}{2} = 4 \\ x + 3 = 8 \\ x = 5 \][/tex]
4. Bottom-right:
[tex]\[ 4 = 4 \][/tex]
This is true, so it is correct.
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ \boxed{x = 5, y = 2} \][/tex]