Problem 69 - Algebra 2

Question

Solve each system of equations by using inverse matrices. $$ \begin{array}{l}{2 x+3 y=8} \\ {x-2 y=-3}\end{array} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 1\) and \(y = 2\).

Step 1: Write the Equations in Matrix Form

Express the given system of equations in the form of a matrix equation. The system of equations:\[\begin{array}{l}2x + 3y = 8 \x - 2y = -3\end{array}\]can be written in matrix form as:\[AX = B\]where \[A = \begin{bmatrix} 2 & 3 \ 1 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \end{bmatrix}, \quad B = \begin{bmatrix} 8 \ -3 \end{bmatrix}\]

Step 2: Find the Inverse of Matrix A

To solve for \(X\), we need the inverse of matrix \(A\). The formula for finding the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is given by:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix:\[A = \begin{bmatrix} 2 & 3 \ 1 & -2 \end{bmatrix}\]we calculate the determinant:\[ad - bc = (2)(-2) - (3)(1) = -4 - 3 = -7\]Then, the inverse is:\[A^{-1} = \frac{1}{-7} \begin{bmatrix} -2 & -3 \ -1 & 2 \end{bmatrix} = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} \ \frac{1}{7} & -\frac{2}{7} \end{bmatrix}\]

Step 3: Solve for X Using the Inverse Matrix

Multiply the inverse matrix \(A^{-1}\) by matrix \(B\) to solve for \(X\):\[X = A^{-1}B\]Substitute the values:\[X = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} \ \frac{1}{7} & -\frac{2}{7} \end{bmatrix} \begin{bmatrix} 8 \ -3 \end{bmatrix}\]Perform the matrix multiplication:\[X = \begin{bmatrix} \frac{2}{7} \times 8 + \frac{3}{7} \times (-3) \ \frac{1}{7} \times 8 + (-\frac{2}{7}) \times (-3) \end{bmatrix} = \begin{bmatrix} \frac{16}{7} - \frac{9}{7} \ \frac{8}{7} + \frac{6}{7} \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}\]

Step 4: Interpret the Solution

The resulting matrix \(X\) from the final multiplication gives us the values of \(x\) and \(y\). Therefore:\[x = 1 \quad \text{and} \quad y = 2\]

Key Concepts

Understanding Matrix Equations

Matrix equations are a powerful way to represent and solve systems of linear equations. Instead of dealing with many equations and variables, we can condense this information into matrices and vectors. Imagine you have a set of equations, like:

• 2x + 3y = 8
• x - 2y = -3

By representing the equations in matrix form, we transform it into: \[AX = B\]

• Matrix \(A\) holds the coefficients of the variables (\(x\) and \(y\))
• Vector \(X\) contains the variables
• Vector \(B\) includes the constants from the equations

Using this unified format makes it easier to apply mathematical operations for solving the equations.

Finding the Inverse of a 2x2 Matrix

The inverse of a matrix is like the reciprocal of a number; when multiplied by the original matrix, it results in the identity matrix, similar to how multiplying a number by its reciprocal results in 1. For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse can be calculated using the formula:\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]where \(ad - bc\) is the determinant of the matrix. An inverse exists only if this determinant is non-zero. In our example, the determinant is \(-7\), so the inverse matrix becomes:\[A^{-1} = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} \ \frac{1}{7} & -\frac{2}{7} \end{bmatrix}\]Finding an inverse is crucial since it's the key to solving the matrix equation \(AX = B\).

Determinant Calculation Simplified

The determinant is a special value calculated from a square matrix, which gives important information about the matrix, such as whether it has an inverse. For a 2x2 matrix, the calculation is straightforward:

• Multiply the top left element (\(a\)) by the bottom right (\(d\))
• Multiply the top right (\(b\)) by the bottom left (\(c\))
• Subtract the second product from the first: \(ad - bc\)

In our system, the determinant of matrix \(A = \begin{bmatrix} 2 & 3 \ 1 & -2 \end{bmatrix}\) is calculated as \((2)(-2) - (3)(1) = -4 - 3 = -7\).This negative value tells us that an inverse does exist, allowing us to proceed with solving the matrix equation.

Solving Systems of Equations with Inverses

Solving a system of linear equations using inverse matrices is efficient and systematic. Once we have the matrix equation \(AX = B\), and the inverse \(A^{-1}\), we solve for \(X\), holding variable solutions.1. Multiply the inverse matrix \(A^{-1}\) by the constant matrix \(B\):\[X = A^{-1}B\]2. Perform the multiplication to find the value of \(X\).In our example, multiplying the inverse matrix by \(B\):\[X = \begin{bmatrix} \frac{2}{7} & \frac{3}{7} \ \frac{1}{7} & -\frac{2}{7} \end{bmatrix} \begin{bmatrix} 8 \ -3 \end{bmatrix} = \begin{bmatrix} 1 \ 2 \end{bmatrix}\]This result provides the solution \(x = 1\) and \(y = 2\), easily interpreting the outcomes for our variables directly from the matrix. This method offers a clear path from equations to solutions, leveraged by matrix operations.