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    Adjoint and Inverse Matrices Linear Equations

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    Geraldine Timpoc
    The document discusses adjoint and inverse matrices. It defines the adjoint of a matrix as the transpose of its cofactor matrix. It provides examples of calculating the adjoint of 2x2 and 3x3 matrices. It then defines the inverse of a matrix and provides the formula for calculating the inverse. Examples are given of calculating the inverse of 2x2 and 3x3 matrices and using the inverse to solve systems of linear equations.

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    Adjoint and Inverse Matrices Linear Equations

    Uploaded by

    Geraldine Timpoc
    57 views9 pages
    The document discusses adjoint and inverse matrices. It defines the adjoint of a matrix as the transpose of its cofactor matrix. It provides examples of calculating the adjoint of 2x2 and 3x3 matrices. It then defines the inverse of a matrix and provides the formula for calculating the inverse. Examples are given of calculating the inverse of 2x2 and 3x3 matrices and using the inverse to solve systems of linear equations.

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    The document discusses adjoint and inverse matrices. It defines the adjoint of a matrix as the transpose of its cofactor matrix. It provides examples of calculating the adjoint of 2x2 and 3x3 matrices. It then defines the inverse of a matrix and provides the formula for calculating the inverse. Examples are given of calculating the inverse of 2x2 and 3x3 matrices and using the inverse to solve systems of linear equations.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    57 views9 pages

    Adjoint and Inverse Matrices Linear Equations

    Uploaded by

    Geraldine Timpoc
    The document discusses adjoint and inverse matrices. It defines the adjoint of a matrix as the transpose of its cofactor matrix. It provides examples of calculating the adjoint of 2x2 and 3x3 matrices. It then defines the inverse of a matrix and provides the formula for calculating the inverse. Examples are given of calculating the inverse of 2x2 and 3x3 matrices and using the inverse to solve systems of linear equations.

    Copyright:

    © All Rights Reserved
    Download as PPTX, PDF, TXT or read online from Scribd
    Download as pptx, pdf, or txt
    of 9

    ADJOINT AND INVERSE MATRICES

    LINEAR EQUATIONS
    Adjoint and Inverse Matrices

    • Adjoint/Adjugate – a transpose of cofactor matrix of the given matrix.


    – denoted by Adj.
    In 2x2 matrix:

    A = 1 2 = a b  Adj.(A) d -b
    -4 -3 c d -c a

    -3 -2
    Adj.(A) =
    4 1
    In 3x3 matrix:
    3 0 2
    A = 2 0 -2 Adj.(A) = ?
    0 1 1
    First: Determine the cofactor of each element.

    3 0 2 3 0 2 3 0 2
    2 0 -2 2 0 -2 2 0 -2
    0 1 1 0 1 1 0 1 1

    1st row 2nd row 3rd row


    -1(0) – 1(-2) = 2 0(1) – 1(2) = -2 0(-2) – 0(2) = 0
    2(1) – 0(-2) = 2 3(1) – 0(2) = 3 3(-2) – 2(2) = -10
    2(1) – 0(0) = 2 3(1) – 0(0) = 3 3(0) – 2(0) = 0 3: Cofactors = (-1)1+1(R + C) x 2(minor) = 2
    • Inverse Matrices – it is the inverse matrix of the given matrix, which when multiplied by each other will result to an identity
    – ex. The inverse of matrix A = A-1.

    Identity: A x A-1 = Inth


    Formula in solving the inverse matrix:

     A-1 =

    *Note: the determinant ≠ 0; if the determinant = 0,


    then the inverse of that matrix does not exist
    In 2x2 matrix:

    1 2 d -b 1 2 -0.6 -0.4 1 0
     A = A-1 = ?  A-1 = AA-1 = I2  =
    -4 -3 -c a -4 -3 0.8 0.2 0 1

    a b -3 -2 1(-0.6)+2(0.8) 1(-0.4)+2(0.2)
     A =  A-1 = AA-1 =
    c d 4 1 -4(-0.6)+(-3)(0.8) -4(-0.4)+(-3)(0.2)

    -0.6 -0.4 1 0
     A-1 = -3 -2 = AA =
    -1

    0.8 0.2 0 1
    4 1
    In 3x3 matrix:
    3 0 2
     A = 2 0 -2 A -1 = ?  A-1 =
    0 1 1

    Determinant = 3(2) – 0(2) + 2(2) = 10

    2 2 0 2 2 0
      Adj.(A) = -2 3 10  A-1 = -2 3 10
    2 -3 0 2 -3 0

    0.2 0.2 0
    A-1 = -0.2 0.3 1
    0.2 -0.3 0
    Linear Equations
    a b
    A=  Adj.(A) = d -b
    c d -c a
    In 2x2:

    2𝑥 − 𝑦=4
     
    l A l: 7 A= 2
    3
    -1
    2
     Adj.(A) = 2
    -3
    1
    2
    A X B

    2 -1 x 4  A-1 = Adj.(A)
    3 2 y 13
     A-1 = 2 1
    -3 2
    AX = B  A-1AX = A-1B
    IX = A-1B X = A-1B: 2x2 2x1 ORDER: 2X1
    X = A-1B
    XA = B  AA-1X = BA-1 x   1 2 1 4
    IX = BA-1 y 7 -3 2 13
    X = BA-1
    x   1 21 3 ∴ x = 3, y = 2
    A B ≠ BA
    -1 -1
    y 7 14 2
    -4 -4 0
    In 3x3:  A = Adj.(A) =
    -1
    -5 1 -3
    x + y – z = -2 3 -3 -3
    2x – y + z = 5 X = A-1B: 3x3 3x1 ORDER: 3X1
    Determinant = -12
    -x + 2y + 2z = 1
    A X B x -4 -4 0 -2
    1 1 -1 x -2   -5 1 -3 5
    y
    2 -1 1 y 5 z 3 -3 -3 1
    -1 2 2 z 1
    x -12
    = 1[2(-1) – 2] – 1[2(2) – 1(-1)] + (-1)[2(2) – (-1)(-1)]  
    y 12
    = 1[-4] – 1[5] + (-1)[3] z 2
    = -4 – 5 – 3 = -12
    x 1
    -4 5 3 -4 -5 3 y -1 ∴ x = 1, y = -1, z = 2
    Adj. (A) = 4 1 3  -4 1 -3 z 2
    0 3 -3 0 -3 -3

    -4 -4 0
    Adj. (A) = -5 1 -3
    3 -3 -3
    THANK YOU FOR LISTENING!

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