8.30 (Algebra: solve linear equations) Write a method that solves the following
2 * 2 system of linear equations:
a 00 x + a 01 y = b 0
a 10 x + a 11 y = b 1
x =
b 0 a 11 - b 1 a 01
a 00 a 11 - a 01 a 10
y =
b 1 a 00 - b 0 a 10
a 00 a 11 - a 01 a 10
The method header is
public static double[] linearEquation(double[][] a, double[] b)
The method returns null if a 00 a 11 - a 01 a 10 is 0 . Write a test program that
prompts the user to enter a 00 , a 01 , a 10 , a 11 , b 0 , and b 1 , and displays the result. If
a 00 a 11 - a 01 a 10 is 0 , report that “The equation has no solution.” A sample run is
similar to Programming Exercise 3.3.
2 * 2 system of linear equations:
a 00 x + a 01 y = b 0
a 10 x + a 11 y = b 1
x =
b 0 a 11 - b 1 a 01
a 00 a 11 - a 01 a 10
y =
b 1 a 00 - b 0 a 10
a 00 a 11 - a 01 a 10
The method header is
public static double[] linearEquation(double[][] a, double[] b)
The method returns null if a 00 a 11 - a 01 a 10 is 0 . Write a test program that
prompts the user to enter a 00 , a 01 , a 10 , a 11 , b 0 , and b 1 , and displays the result. If
a 00 a 11 - a 01 a 10 is 0 , report that “The equation has no solution.” A sample run is
similar to Programming Exercise 3.3.
package Chapter_08; import java.util.Scanner; public class Exercise_30 { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.println("Enter a, b, c, d, e, f below:"); double[][] a = new double[2][2]; double[] b = new double[2]; for (int i = 0; i < a.length; i++) for (int j = 0; j < a[i].length; i++) a[i][j] = input.nextDouble(); double[] point = linearEquation(a, b); if (linearEquation(a, b) == null) { System.out.println("No solution."); } else { System.out.println("x = " + point[0] + " and y = " + point[1]); } } public static double[] linearEquation(double[][] a, double[] b) { // equation: // ab_Minus_bc = a * d - b * c double ab_Minus_bc = a[0][0] * a[1][1] - a[0][1] * a[1][0]; if (ab_Minus_bc == 0.0) return null; // equation: // x = (e * d - b * f) / ab_Minus_bc; // y = (a * f - e * c) / ab_Minus_bc; double[] point = new double[2]; point[0] = (b[0] * a[1][1] - b[1] * a[0][1]) / ab_Minus_bc; point[1] = (b[1] * a[0][0] - b[0] * a[1][0]) / ab_Minus_bc; return point; } }
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