Problem 48 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use an inverse matrix to solve (if possible) the system of linear equations. $$\left\\{\begin{array}{c} 0.2 x-0.6 y=2.4 \\ -x+1.4 y=-8.8 \end{array}\right.$$

Step-by-Step Solution

Answer

The system of equations is solved as x = 10 and y = -8.

Step 1: Represent the system of equations as a matrix

The system can be written in the form of \(AX = B\), where A is the matrix of coefficients of the variables, X is the column matrix of variables [x, y] and B is the column matrix of the constants on the right-hand side of the equations. So, \(A=\left[\begin{array}{cc}0.2 & -0.6 \\ -1 & 1.4 \end{array}\right]\), \(X=\left[\begin{array}{c}x \\ y \end{array}\right]\) and \(B=\left[\begin{array}{c}2.4 \\ -8.8 \end{array}\right]\).

Step 2: Calculate the inverse of matrix A

The inverse of a 2x2 matrix \(\left[\begin{array}{cc}a & b \\ c & d \end{array}\right]\) is given by \(1/\text{det}(A)\times\left[\begin{array}{cc}d & -b \\ -c & a \end{array}\right]\) where det(A) is the determinant of the matrix A, calculated as \(ad− bc\). Applying this to matrix A, we get \(A^{-1}= 1/-1.44\times\left[\begin{array}{cc}1.4 & 0.6 \\ 1 & 0.2 \end{array}\right]= \left[\begin{array}{cc}-0.97222 & -0.41667 \\ -0.69444 & -0.13889 \end{array}\right]\)

Step 3: Multiply the inverse matrix with matrix B

The solution to the system of equations is given by \(X = A^{-1}B\). Multiplying the inverse matrix A calculated in step 2 with matrix B, we get \(X=\left[\begin{array}{cc}-0.97222 & -0.41667 \\ -0.69444 & -0.13889 \end{array}\right]\times\left[\begin{array}{c}2.4 \\ -8.8 \end{array}\right]=\left[\begin{array}{c}x \\ y \end{array}\right]=\left[\begin{array}{c}10 \\ -8 \end{array}\right]\), indicating that the solution to the system of equations is \(x=10\) and \(y=-8\).

Key Concepts

System of Linear Equations

Linear equations involve variables like \(x\) and \(y\) along with constants. They are called 'linear' because the graph of such equations forms a straight line. A system of linear equations is when we have more than one linear equation working together. In our exercise, the system is:

• \(0.2x - 0.6y = 2.4\)
• \(-x + 1.4y = -8.8\)

This means we're looking for values of \(x\) and \(y\) that satisfy both equations at the same time. Solving systems of linear equations means finding a point, an \(x\) and \(y\), where these lines intersect, indicating the solution.

Matrix Representation

Representing a system of equations with matrices is a neat and organized way to handle calculations. We express our system in the form \(AX = B\). Here:

• \(A\) is the coefficients matrix, containing the coefficients of the variables from both equations. For our case: \[A = \begin{bmatrix} 0.2 & -0.6 \ -1 & 1.4 \end{bmatrix}\]
• \(X\) is the column matrix, a vector of variables: \[X = \begin{bmatrix} x \ y \end{bmatrix}\]
• \(B\) is the constants matrix from the right-hand side: \[B = \begin{bmatrix} 2.4 \ -8.8 \end{bmatrix}\]

Using this representation makes the process of finding \(x\) and \(y\) much clearer, especially when applying matrix operations.

Determinant Calculation

The determinant of a matrix is a special number that can be calculated from its elements. It provides crucial information about the matrix, such as whether it's invertible. For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant \(\text{det}(A)\) is computed as \(ad - bc\).In our exercise, for matrix \(A\), which is:\[A = \begin{bmatrix} 0.2 & -0.6 \ -1 & 1.4 \end{bmatrix}\]the determinant is calculated as:\[(0.2)(1.4) - (-0.6)(-1) = 0.28 - 0.6 = -0.32\]A non-zero determinant means the matrix is invertible, which is essential for solving the system using the inverse matrix method.

Multiplication of Matrices

Matrix multiplication involves taking two matrices and producing a new matrix. It is not quite the same as multiplying numbers; instead, it includes summing up products of rows and columns. When calculating \(A^{-1}B\), we find the solution to the linear equations.Given the inverse matrix \[A^{-1} = \begin{bmatrix} -0.97222 & -0.41667 \ -0.69444 & -0.13889 \end{bmatrix}\]and matrix \(B\), which is\[B = \begin{bmatrix} 2.4 \ -8.8 \end{bmatrix}\],the multiplication is done as follows:

• First row of \(A^{-1}\) dot first column of \(B\) gives the first element of matrix \(X\): \[(-0.97222 \times 2.4) + (-0.41667 \times -8.8) = 10\]
• Second row of \(A^{-1}\) dot first column of \(B\) provides the second element:\[(-0.69444 \times 2.4) + (-0.13889 \times -8.8) = -8\]

Thus, we find \(x = 10\) and \(y = -8\), effectively solving the system of equations by means of matrix operations.