Answer :
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]