Problem 47 - A Graphical Approach to College Algebra

Question

Solve each system by using the matrix inverse method. $$\begin{array}{r} x-\sqrt{2} y=2.6 \\ 0.75 x+y=-7 \end{array}$$

Step-by-Step Solution

Answer

Use the matrix inverse method to find \(x\) and \(y\).

Step 1: Write the System as a Matrix Equation

First, write the given system of equations as a matrix equation. The system \( \begin{align*} x - \sqrt{2} y &= 2.6 \ 0.75x + y &= -7 \end{align*} \) can be represented in matrix form as: \[ \begin{bmatrix} 1 & -\sqrt{2} \ 0.75 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 2.6 \ -7 \end{bmatrix} \].

Step 2: Find the Inverse of the Coefficient Matrix

To solve for \( \begin{bmatrix} x \ y \end{bmatrix} \), find the inverse of the coefficient matrix \( A = \begin{bmatrix} 1 & -\sqrt{2} \ 0.75 & 1 \end{bmatrix} \). The formula to find the inverse of a \(2 \times 2\) matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is:\[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \].For \( A \), we have:\[ ad - bc = 1 \cdot 1 - (-\sqrt{2}) \cdot 0.75 = 1 + 0.75\sqrt{2} \].

Step 3: Calculate the Determinant

The determinant (\(ad - bc\)) is \(1 + 0.75\sqrt{2}\). Compute this expression to confirm that the determinant is non-zero, meaning the inverse exists.

Step 4: Compute the Inverse

Using the determinant from Step 3, calculate the inverse of the matrix \( A \). Place each element as follows:\[ A^{-1} = \frac{1}{1 + 0.75\sqrt{2}} \begin{bmatrix} 1 & \sqrt{2} \ -0.75 & 1 \end{bmatrix} \].Ensure to simplify each element within the matrix.

Step 5: Solve for the Variables using the Inverse

Now, use the matrix inverse to solve for \( \begin{bmatrix} x \ y \end{bmatrix} \): \[ \begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \begin{bmatrix} 2.6 \ -7 \end{bmatrix} \]. Substitute \( A^{-1} \) and compute the resulting matrix multiplication.

Step 6: Compute the Values of \(x\) and \(y\)

Perform the matrix multiplication from Step 5:\[ \begin{bmatrix} x \ y \end{bmatrix} = \frac{1}{1 + 0.75\sqrt{2}} \begin{bmatrix} 1 & \sqrt{2} \ -0.75 & 1 \end{bmatrix} \begin{bmatrix} 2.6 \ -7 \end{bmatrix} \].Calculate each element separately to find the values of \(x\) and \(y\).

Step 7: State the Solution

After performing the multiplication and any necessary simplifications, you'll find the values of \(x\) and \(y\) that satisfy both equations.

Key Concepts

Systems of Equations

When you encounter a system of equations like \[ \begin{align*} x - \sqrt{2} y &= 2.6 \ 0.75x + y &= -7 \end{align*} \]it means you're being asked to find the values of the variables, in this case \(x\) and \(y\), that make both equations true simultaneously. Systems of equations can have:

• One unique solution, where the lines intersect at one point.
• No solution, where the lines are parallel.
• Infinitely many solutions, where the lines coincide.

In this exercise, we employ the **Matrix Inverse Method**. This method is only applicable when there is a unique solution, as it relies on the existence of an inverse for the coefficient matrix. Initially, systems of equations can seem daunting, but the Matrix Inverse Method simplifies solving by converting the system into a manageable matrix equation.

Determinant Calculation

In the context of solving equations with matrices, the determinant is a crucial element. It is a special value that can be calculated from a square matrix. For a \(2 \times 2\) matrix, say \[ \begin{bmatrix} a & b \ c & d \end{bmatrix} \]the determinant is calculated using the formula:\[ ad - bc \]This determinant must be non-zero to use the matrix inverse method. If the determinant is zero, the matrix does not have an inverse, indicating the system does not have a unique solution.

In the given exercise, the determinant is calculated for the matrix \[ \begin{bmatrix} 1 & -\sqrt{2} \ 0.75 & 1 \end{bmatrix} \], yielding: \[ 1 \cdot 1 - (-\sqrt{2}) \cdot 0.75 = 1 + 0.75\sqrt{2} \]This is a positive, non-zero value, confirming that our matrix has an inverse and the system has a unique solution. Understanding how to compute and interpret the determinant is vital for verifying the solvability of systems using matrices.

Matrix Algebra

Matrices can seem intimidating at first with their rows and columns, but they are powerful tools in algebra and beyond. In the context of solving equations, we use several operations in matrix algebra:

• **Matrix multiplication**: When we multiply matrices, we're performing operations row by column. It's essential to align the matrix dimensions correctly to ensure the multiplication is valid.
• **Matrix inversion**: This involves finding a matrix that, when multiplied with the original matrix, results in the identity matrix. For the \(2 \times 2\) coefficient matrix \(A\) from the exercise, the inverse \(A^{-1}\) is calculated as:
  \[ A^{-1} = \frac{1}{1 + 0.75\sqrt{2}} \begin{bmatrix} 1 & \sqrt{2} \ -0.75 & 1 \end{bmatrix} \]

Once you have the inverse, solving the equation \[ \begin{bmatrix} x \ y \end{bmatrix} = A^{-1} \begin{bmatrix} 2.6 \ -7 \end{bmatrix} \]involves simply performing the matrix multiplication. Matrix algebra allows us to handle complex problems efficiently, making it a crucial element in modern mathematical toolkits.