Question

Solve each set of equations by the method of finding the inverse of the coefficient matrix. $$ \left\\{\begin{array}{c} x-2 y=5 \\ 3 x+y=15 \end{array}\right. $$

Short answer

The solution is \( x = 5 \) and \( y = 0 \).

Step-by-step solution

Write the system of equations in matrix form

Express the given system of equations \[ \begin{cases} x-2y=5 \ 3x+y=15 \end{cases} \] in the matrix form \( A\boldsymbol{x} = \boldsymbol{b} \) where \[A = \begin{pmatrix} 1 & -2 \ 3 & 1 \end{pmatrix}, \boldsymbol{x} = \begin{pmatrix} x \ y \end{pmatrix}, \boldsymbol{b} = \begin{pmatrix} 5 \ 15 \end{pmatrix} \]

Find the inverse of matrix A

The inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \) is given by \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \] Applying this formula, \[ A^{-1} = \frac{1}{(1)(1) - (-2)(3)} \begin{pmatrix} 1 & 2 \ -3 & 1 \end{pmatrix} = \frac{1}{1 + 6} \begin{pmatrix} 1 & 2 \ -3 & 1 \end{pmatrix} = \frac{1}{7} \begin{pmatrix} 1 & 2 \ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \ -\frac{3}{7} & \frac{1}{7} \end{pmatrix} \]

Multiply both sides by the inverse matrix

To find \( \boldsymbol{x} \), we multiply both sides of the equation \( A\boldsymbol{x} = \boldsymbol{b} \) by \( A^{-1} \). \[ \boldsymbol{x} = A^{-1} \boldsymbol{b} \] Substituting \( A^{-1} \) and \( \boldsymbol{b} \): \[ \boldsymbol{x} = \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \ -\frac{3}{7} & \frac{1}{7} \end{pmatrix} \begin{pmatrix} 5 \ 15 \end{pmatrix} \] Perform the matrix multiplication: \[ \boldsymbol{x} = \begin{pmatrix} \frac{1}{7}(5) + \frac{2}{7}(15) \ -\frac{3}{7}(5) + \frac{1}{7}(15) \end{pmatrix} = \begin{pmatrix} \frac{5}{7} + \frac{30}{7} \ -\frac{15}{7} + \frac{15}{7} \end{pmatrix} = \begin{pmatrix} \frac{35}{7} \ 0 \end{pmatrix} = \begin{pmatrix} 5 \ 0 \end{pmatrix} \]

Interpret the solution

The solution to the system of equations is \( x = 5 \) and \( y = 0 \).

Key concepts

inverse matrix

An inverse matrix is like the opposite of a matrix. If we have a matrix A, its inverse, written as A^{-1}, undoes what the original matrix does. When you multiply a matrix by its inverse, you get the identity matrix, which acts like the number 1 in regular multiplication. To find the inverse of a 2x2 matrix, you use the formula: \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} \] Here, a, b, c, and d are the elements of the matrix in this arrangement: \[ A = \begin{pmatrix} a & b \ c & d \end{pmatrix} \] It’s important that the determinant (ad - bc) is not zero because you can’t divide by zero! The determinant helps you understand if the matrix can be inverted.

matrix form of linear equations

In solving systems of linear equations, we often rewrite the equations using matrices. For our example:\[ \begin{cases} x - 2y = 5 \ 3x + y = 15 \end{cases} \] This can be written in matrix form as A **x** = **b** where:\[ A = \begin{pmatrix} 1 & -2 \ 3 & 1 \end{pmatrix}, \quad \boldsymbol{x} = \begin{pmatrix} x \ y \end{pmatrix}, \quad \text{and} \quad \boldsymbol{b} = \begin{pmatrix} 5 \ 15 \end{pmatrix} \] Here, A is called the coefficient matrix containing the coefficients from your equations, **x** is the column vector containing your unknowns (x, y), and **b** is the column vector containing the constants from the right side of the equations. This compact matrix form makes it easier to apply mathematical operations to solve the system.

matrix multiplication

Matrix multiplication is an essential operation while solving systems of equations with matrices. For our example: \[ \boldsymbol{x} = A^{-1} \boldsymbol{b} \]To compute this, we multiply the inverse of matrix A by the vector **b**. Here’s how we multiply:\[ \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \ -\frac{3}{7} & \frac{1}{7} \end{pmatrix} \begin{pmatrix} 5 \ 15 \end{pmatrix} \]To perform the multiplication, we do the following:- Multiply the first row of A^{-1} by the elements of **b**- Multiply the second row of A^{-1} by the elements of **b**We get:\[ \begin{pmatrix} \frac{1}{7}(5) + \frac{2}{7}(15) \ -\frac{3}{7}(5) + \frac{1}{7}(15) \end{pmatrix} = \begin{pmatrix} 5 \ 0 \end{pmatrix} \]Notice how each element from the resulting matrix comes from the combination of the rows and columns from the matrices involved.

2x2 matrices

2x2 matrices are among the simplest and most intuitive types of matrices. They contain four elements arranged in two rows and two columns. For example:\[ \begin{pmatrix} a & b \ c & d \end{pmatrix} \]These matrices are important in many branches of mathematics and offer a great way to learn basic matrix operations, like addition, subtraction, and importantly, inversion and multiplication. In our exercise:\[ A = \begin{pmatrix} 1 & -2 \ 3 & 1 \end{pmatrix} \]We see how a 2x2 matrix can be used to set up a system of linear equations, compute its inverse, and then solve it. Working with 2x2 matrices lays the foundation for understanding larger matrices and more complex mathematical concepts.