Linear Algebra & Trigonometry Study Notes

LINEAR ALGEBRA & TRIGONOMETRY Notes

SYSTEM OF EQUATIONS

  • Problem Statement: Solve the following system of equations:

    • Equation 1: x+2y+z=3x + 2y + z = 3

    • Equation 2: x+3y+7z=1x + 3y + 7z = -1

    • Equation 3: 3y2z=73y - 2z = 7

Method: Cramer's Rule
  • Cramer's Rule is a mathematical theorem used to solve systems of linear equations with as many equations as unknowns, using determinants.

  • Definitions:

    • Let the system of equations be represented in matrix form as:
      AX=BAX = B where:

    • A is the coefficient matrix

    • X is the column matrix of variables

    • B is the column matrix of the constants.

MATRIX OPERATIONS

  • Matrix Y:

    • Matrix Y is represented as:
      (15amp;9 3amp;0 6amp;1 8amp;11 9amp;10 9amp;5 14amp;20 13amp;7 6amp;3 4amp;18 16amp;10 155 921  )\begin{pmatrix} 15 & 9 \ 3 & 0 \ -6 & 1 \ 8 & 11 \ 9 & 10 \ -9 & 5 \ 14 & 20 \ 13 & 7 \ 6 & 3 \ 4 & 18 \ 16 & 10 \ 155 \ 921 \ \ \end{pmatrix}

Product of Matrices
  • Finding the Product: To find the product of two matrices, the inner dimensions must match; if matrix A is of size mimesnm imes n and matrix B is of size nimespn imes p, the resulting matrix will be of size mimespm imes p.

  • The product can be calculated as:
    Cij=extdotproductoftheithrowofAandthejthcolumnofBC_{ij} = ext{dot product of the i-th row of A and the j-th column of B}

MATRIX INVERSION

  • Problem Statement: Find the inverse of the matrix:
    (4amp;18 16amp;10 155 921  )\begin{pmatrix} 4 & 18 \ 16 & 10 \ 155 \ 921 \ \ \end{pmatrix}

Inverse Matrix Definition:
  • The inverse of a matrix A, denoted as A1A^{-1}, is a matrix such that:
    AA1=A1A=IAA^{-1} = A^{-1}A = I
    Where I is the identity matrix. Not all matrices have inverses (only non-singular matrices do). Chek the determinant to ensure it is non-zero before attempting to find the inverse.

Process to Find Inverse:
  1. Calculate the determinant of the matrix.

  2. If the determinant is non-zero, use the formula to find the inverse:
    A1=rac1extdet(A)extadj(A)A^{-1} = rac{1}{ ext{det}(A)} ext{adj}(A) where adj(A) is the adjugate of A.