Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.