Matrix and Determinant - Question 3 (On System of Linear Equations)

• We will now solve a question on system of linear equations using matrices and determinants. Given equation:
• 2x + 3y = 12
• 4x - 5y = 7

Solution Steps:

1. Write the given system of linear equations in matrix form: | 2 3 | | x | | 12 | | 4 -5 | | y | = | 7 |

2. Write the coefficient matrix, variable matrix, and constant matrix: Coefficient matrix (A): | 2 3 | | 4 -5 | Variable matrix (X): | x | | y | Constant matrix (B): | 12 | | 7 |

3. Calculate the determinant of the coefficient matrix (A): |A| = 2*(-5) - 3*4 = -10 - 12 = -22

4. If |A| ≠ 0, the system has a unique solution. If |A| = 0, the system may have either no solution or infinite solutions.

5. Find the inverse of the coefficient matrix (A): A^-1 = 1/|A| * | -5 -3 | | -4 2 |

6. Multiply A^-1 with the constant matrix (B): A^-1 * B = | -5 -3 | * | 12 | = | x | | -4 2 | | y |

7. Solve for x and y: -5x - 3y = 12 -4x + 2y = 7

8. Simplify the equations: -5x - 3y = 12 –> -5x = 12 + 3y –> x = (-12 - 3y) / 5 (1) -4x + 2y = 7 –> -4x = -2y + 7 –> x = (-2y + 7) / 4 (2)

9. Equate the values of x from equations (1) and (2): (-12 - 3y) / 5 = (-2y + 7) / 4

10. Solve for y and substitute the value of y in any one of the equations to find x.

Matrix and Determinant - Question 3 (On System of Linear Equations) - Solution Steps

Slide 11:

• We have the following system of linear equations:
  • 2x + 3y = 12
  • 4x - 5y = 7

Slide 12:

• Writing the equations in matrix form:
  • | 2 3 | | x | | 12 |
  • | 4 -5 | | y | = | 7 |

Slide 13:

• Coefficient matrix (A):

  • | 2 3 |
  • | 4 -5 |

• Variable matrix (X):

  • | x |
  • | y |

Slide 14:

• Constant matrix (B):
  • | 12 |
  • | 7 |

Slide 15:

• Calculate the determinant of the coefficient matrix (A):
  • |A| = 2*(-5) - 3*4 = -10 - 12 = -22

Slide 16:

• If |A| ≠ 0, the system has a unique solution.

Slide 17:

• Find the inverse of the coefficient matrix (A):
  • A^-1 = 1/|A| * | -5 -3 | | -4 2 |

Slide 18:

• Multiply A^-1 with the constant matrix (B):
  • A^-1 * B = | -5 -3 | * | 12 | = | x | | -4 2 | | y |

Slide 19:

• Simplify the equations obtained:
  • -5x - 3y = 12
  • -4x + 2y = 7

Slide 20:

• Solve for x and y:
  • Use algebraic methods to find the values of x and y

Slide 21:

• Let’s solve the equation obtained in the previous step:
  • -5x - 3y = 12 (Equation 1)
  • -4x + 2y = 7 (Equation 2)

Slide 22:

• Simplifying Equation 1, we get:
  • -5x = 12 + 3y

Slide 23:

• Solving Equation 1 for x, we get:
  • x = (-12 - 3y) / 5

Slide 24:

• By rearranging Equation 2, we find:
  • -4x = -2y + 7

Slide 25:

• Solving Equation 2 for x, we get:
  • x = (-2y + 7) / 4

Slide 26:

• Equating the values of x from Equations 1 and 2, we get:
  • (-12 - 3y) / 5 = (-2y + 7) / 4

Slide 27:

• Cross multiplying and simplifying the equation, we get:
  • 4(-12 - 3y) = 5(-2y + 7)

Slide 28:

• Expanding and simplifying the equation further, we get:
  • -48 - 12y = -10y + 35

Slide 29:

• Rearranging and solving for y, we get:
  • y = 23/22

Slide 30:

• Substitute the value of y in any one of the equations to find x.