Problem 42 - Algebra and Trigonometry

Question

Solve for \(x\) and \(y\). $$ \left[\begin{array}{rr}{x} & {y} \\ {-y} & {x}\end{array}\right]-\left[\begin{array}{rr}{y} & {x} \\ {x} & {-y}\end{array}\right]=\left[\begin{array}{rr}{4} & {-4} \\ {-6} & {6}\end{array}\right] $$

Step-by-Step Solution

Answer

\(x = 5\), \(y = 1\)

Step 1: Understand the Matrix Equation

We are given a matrix equation \(\left[\begin{array}{rr}{x} & {y} \ {-y} & {x}\end{array}\right]-\left[\begin{array}{rr}{y} & {x} \ {x} &{-y}\end{array}\right]=\left[\begin{array}{rr}{4} & {-4} \ {-6} &{6}\end{array}\right]\). This means we need to subtract the matrices on the left and set them equal to the matrix on the right.

Step 2: Subtract the Matrices

Subtract the corresponding elements of the matrices on the left side of the equation. The result is another matrix: \(\left[\begin{array}{rr}{x-y} & {y-x} \ {-y-x} & {x+y}\end{array}\right]\).

Step 3: Set Up Equations from the Resulting Matrix

Equate each element of the resulting matrix to the corresponding element of the matrix on the right side of the equation:1. \(x-y = 4\)2. \(y-x = -4\)3. \(-y-x = -6\)4. \(x+y = 6\)

Step 4: Solve the First Pair of Equations

Solve the first two equations:1. \(x - y = 4\) 2. \(y - x = -4\)Adding these equations, \(x - y + y - x = 4 - 4\). Simplifies to \(0 = 0\), which is redundant. So move on to a systematic approach for checking consistency with additional equations.

Step 5: Solve the Third Equation

Use \(-y-x = -6\). Rearrange to find \(x + y = 6\), which is consistent with equation 4.

Step 6: Solve for \(x\) and \(y\) Using Consistent Equations

We have \(x-y = 4\) and \(x+y = 6\). Add these equations to eliminate \(y\):\((x-y) + (x+y) = 4 + 6\) leads to \(2x = 10\), so \(x = 5\).Substitute \(x = 5\) back into \(x+y = 6\), so \(5 + y = 6\). Hence, \(y = 1\).

Step 7: Verify Solution

Substitute \(x = 5\) and \(y = 1\) back into the original matrix equation to ensure the left-hand side equals the right-hand side:\(\left[\begin{array}{rr}{5} & {1} \ {-1} & {5}\end{array}\right]-\left[\begin{array}{rr}{1} & {5} \ {5} &{-1}\end{array}\right]=\left[\begin{array}{rr}{4} & {-4} \ {-6} &{6}\end{array}\right]\)This verifies that the solution \(x = 5\) and \(y = 1\) is correct.

Key Concepts

Matrix Subtraction

Matrix subtraction is similar to subtracting numbers, but it happens element by element. Just like with numbers, you align the matrices and subtract each element in the corresponding position. To perform matrix subtraction, the matrices involved must be of the same size. This ensures that every element has a direct counterpart in the other matrix to subtract from or be subtracted.

In the exercise, we need to subtract two 2x2 matrices. We do this by taking each element in the first matrix and subtracting the element in the same position in the second matrix. For example, if we have two matrices:

• Matrix A: \left[\, \begin{array}{cc} a & b \ c & d \end{array} \, \right]
• Matrix B: \left[\, \begin{array}{cc} e & f \ g & h \end{array} \, \right]

Subtracting B from A gives us a resulting matrix where each element is calculated as:

• \(a - e\)
• \(b - f\)
• \(c - g\)
• \(d - h\)

By following this procedure, you will get a new matrix that represents the difference between the two matrices involved.

Solving Linear Equations

Solving linear equations involves finding the values of variables that satisfy the equation. In our scenario, we have equations arising from comparing elements of the resulting matrix of our matrix subtraction with those of the matrix on the right of the equation.

A systematic way to solve linear equations is by first aligning and simplifying the equations as much as possible. Then, using methods like substitution or elimination can help resolve the equations into a single solution. In our exercise, these compared equations form a system:

• \(x - y = 4\)
• \(y - x = -4\)
• \(-y - x = -6\)
• \(x + y = 6\)

From this, we can directly solve some equations or add them together to eliminate variables, making it easier to solve for one variable first. This calculated value is then substituted back into one of the original equations to solve for the second variable.

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. The goal is to find values for these variables that satisfy all equations in the system simultaneously. In our exercise, after performing matrix subtraction, we compare coefficients to form our system of equations.

There are several methods to solve a system of equations:

• The **graphical method**, which isn't practical in this case, but it involves plotting the equations and finding their intersection point.
• The **substitution method**, where one of the variables is expressed in terms of the other, and this is then substituted into another equation.
• The **elimination method**, which involves adding or subtracting equations to eliminate one of the variables, simplifying the system into a single equation in one variable.

In our exercise, we used a combination of all of these principles: systematically eliminating variables to simplify the solution. This approach leads us to the correct solution for the variables involved, ensuring the system's consistency is maintained throughout.