Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Sure, let’s go through the solution step-by-step.
We are given the following matrix equation:
[tex]\[ \left[\begin{array}{cc}x+y & 2 \\ -2 & -y+x\end{array}\right] + \left[\begin{array}{cc}-2 & -1 \\ 3 & -4\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] \][/tex]
First, add the corresponding elements of the two matrices on the left-hand side to simplify the equation.
### Step 1: Calculate the elements of the resulting matrix
1. For the element at position (1, 1):
[tex]\[ (x+y) + (-2) = x + y - 2 \][/tex]
2. For the element at position (1, 2):
[tex]\[ 2 + (-1) = 1 \][/tex]
3. For the element at position (2, 1):
[tex]\[ -2 + 3 = 1 \][/tex]
4. For the element at position (2, 2):
[tex]\[ (-y + x) + (-4) = -y + x - 4 \][/tex]
### Step 2: Form the resulting matrix
After adding the elements, the resulting matrix is:
[tex]\[ \left[\begin{array}{cc} x + y - 2 & 1 \\ 1 & -y + x - 4 \end{array}\right] \][/tex]
### Step 3: Set up the equations
We compare each element of the resulting matrix to the corresponding element in the given matrix on the right-hand side:
[tex]\[ \left[\begin{array}{cc} x + y - 2 & 1 \\ 1 & -y + x - 4 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \][/tex]
From this comparison, we obtain the following system of linear equations:
1. [tex]\(x + y - 2 = 0\)[/tex]
2. [tex]\(1 = 1\)[/tex] (This is always true and doesn't provide additional information)
3. [tex]\(1 = 1\)[/tex] (This is also always true and doesn't provide additional information)
4. [tex]\(-y + x - 4 = 0\)[/tex]
So, we are left with two meaningful equations:
[tex]\[ \begin{cases} x + y - 2 = 0 \\ -x + y - 4 = 0 \end{cases} \][/tex]
### Step 4: Solve the system of equations
Now, solve these simultaneous equations:
1. From [tex]\(x + y - 2 = 0\)[/tex]:
[tex]\[ x + y = 2 \quad \text{(Equation 1)} \][/tex]
2. From [tex]\(-y + x - 4 = 0\)[/tex]:
[tex]\[ -y + x = 4 \quad \text{(Equation 2)} \][/tex]
To solve this system, add Equation 1 and Equation 2:
[tex]\[ (x + y) + (x - y) = 2 + 4 \][/tex]
This simplifies to:
[tex]\[ 2x = 6 \][/tex]
So:
[tex]\[ x = 3 \][/tex]
Now, substitute [tex]\(x = 3\)[/tex] back into Equation 1:
[tex]\[ 3 + y = 2 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = 2 - 3 \][/tex]
[tex]\[ y = -1 \][/tex]
### Step 5: Verify the solution
Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = -1\)[/tex] back into the original equations to confirm they hold true.
1. [tex]\(x + y = 3 + (-1) = 2\)[/tex]
2. [tex]\(-y + x = -(-1) + 3 = 1 + 3 = 4\)[/tex]
Both equations are satisfied.
Hence, the solution is:
[tex]\[ x = 3 \quad \text{and} \quad y = -1 \][/tex]
We are given the following matrix equation:
[tex]\[ \left[\begin{array}{cc}x+y & 2 \\ -2 & -y+x\end{array}\right] + \left[\begin{array}{cc}-2 & -1 \\ 3 & -4\end{array}\right] = \left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right] \][/tex]
First, add the corresponding elements of the two matrices on the left-hand side to simplify the equation.
### Step 1: Calculate the elements of the resulting matrix
1. For the element at position (1, 1):
[tex]\[ (x+y) + (-2) = x + y - 2 \][/tex]
2. For the element at position (1, 2):
[tex]\[ 2 + (-1) = 1 \][/tex]
3. For the element at position (2, 1):
[tex]\[ -2 + 3 = 1 \][/tex]
4. For the element at position (2, 2):
[tex]\[ (-y + x) + (-4) = -y + x - 4 \][/tex]
### Step 2: Form the resulting matrix
After adding the elements, the resulting matrix is:
[tex]\[ \left[\begin{array}{cc} x + y - 2 & 1 \\ 1 & -y + x - 4 \end{array}\right] \][/tex]
### Step 3: Set up the equations
We compare each element of the resulting matrix to the corresponding element in the given matrix on the right-hand side:
[tex]\[ \left[\begin{array}{cc} x + y - 2 & 1 \\ 1 & -y + x - 4 \end{array}\right] = \left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right] \][/tex]
From this comparison, we obtain the following system of linear equations:
1. [tex]\(x + y - 2 = 0\)[/tex]
2. [tex]\(1 = 1\)[/tex] (This is always true and doesn't provide additional information)
3. [tex]\(1 = 1\)[/tex] (This is also always true and doesn't provide additional information)
4. [tex]\(-y + x - 4 = 0\)[/tex]
So, we are left with two meaningful equations:
[tex]\[ \begin{cases} x + y - 2 = 0 \\ -x + y - 4 = 0 \end{cases} \][/tex]
### Step 4: Solve the system of equations
Now, solve these simultaneous equations:
1. From [tex]\(x + y - 2 = 0\)[/tex]:
[tex]\[ x + y = 2 \quad \text{(Equation 1)} \][/tex]
2. From [tex]\(-y + x - 4 = 0\)[/tex]:
[tex]\[ -y + x = 4 \quad \text{(Equation 2)} \][/tex]
To solve this system, add Equation 1 and Equation 2:
[tex]\[ (x + y) + (x - y) = 2 + 4 \][/tex]
This simplifies to:
[tex]\[ 2x = 6 \][/tex]
So:
[tex]\[ x = 3 \][/tex]
Now, substitute [tex]\(x = 3\)[/tex] back into Equation 1:
[tex]\[ 3 + y = 2 \][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[ y = 2 - 3 \][/tex]
[tex]\[ y = -1 \][/tex]
### Step 5: Verify the solution
Substitute [tex]\(x = 3\)[/tex] and [tex]\(y = -1\)[/tex] back into the original equations to confirm they hold true.
1. [tex]\(x + y = 3 + (-1) = 2\)[/tex]
2. [tex]\(-y + x = -(-1) + 3 = 1 + 3 = 4\)[/tex]
Both equations are satisfied.
Hence, the solution is:
[tex]\[ x = 3 \quad \text{and} \quad y = -1 \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
