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    Matrices and Determinant Set 2

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    Jeetamitra Nayak
    This document provides examples and solutions to problems involving matrices and determinants. 1) It gives an example of matrices A and B where AB = BA = O but A ≠ 0 and B ≠ 0. 2) It finds the possible values of x and y given the equality of two determinants. 3) It finds the determinant of a square matrix A given that A adj A = |A|I.

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    Matrices and Determinant Set 2

    Uploaded by

    Jeetamitra Nayak
    19 views3 pages
    This document provides examples and solutions to problems involving matrices and determinants. 1) It gives an example of matrices A and B where AB = BA = O but A ≠ 0 and B ≠ 0. 2) It finds the possible values of x and y given the equality of two determinants. 3) It finds the determinant of a square matrix A given that A adj A = |A|I.

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    Matrices and determinant set 2

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    This document provides examples and solutions to problems involving matrices and determinants. 1) It gives an example of matrices A and B where AB = BA = O but A ≠ 0 and B ≠ 0. 2) It finds the possible values of x and y given the equality of two determinants. 3) It finds the determinant of a square matrix A given that A adj A = |A|I.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    19 views3 pages

    Matrices and Determinant Set 2

    Uploaded by

    Jeetamitra Nayak
    This document provides examples and solutions to problems involving matrices and determinants. 1) It gives an example of matrices A and B where AB = BA = O but A ≠ 0 and B ≠ 0. 2) It finds the possible values of x and y given the equality of two determinants. 3) It finds the determinant of a square matrix A given that A adj A = |A|I.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    Set 2

    Matrices and Determinants MM 15

    1. Give example of matrices A and B such that AB=O = BA 𝐴 ≠ 0, 𝐵 ≠ 0 (1)


    1 0 0 0
    Solution 1 . A = [ ],B=[ ] ⇒ 𝐴𝐵 = 𝑂 = 𝐵𝐴
    0 0 0 1
    3 𝑦 3 2
    2. If | |=| | ,find the possible values of x and y and x,y∈ 𝑁 (1)
    𝑥 1 4 1
    Solution 2. 3× 1 − 𝑥 × 𝑦 = 3 × 1 − 4 × 2
    ⇒ 𝑥𝑦 = 8 ⇒ 𝑥 = 1, 𝑦 = 8; 𝑥 = 2, 𝑦 = 4; 𝑥 = 4, 𝑦 = 2; 𝑥 = 8, 𝑦 = 1
    8 0
    3. If A is a square matrix of order 2 , A. adj A= [ ], find |𝐴| (1)
    0 8
    Solution 3. As A is a square matrix of order 2
    8 0
    A.adjA= |𝐴|𝐼=[ ] ⇒ |𝐴|𝐼 = 8𝐼 ⇒ |𝐴| = 8
    0 8

    For questions 4 and 5


    Statement I is called Assertion (A) and Statement II is called reason (R).
    Read the given statements carefully and choose the correct answer from the four
    options .
    (a) Both the statements are true and Statement II is a correct explanation of I
    (b) Both the statements are true and Statement II is not the correct explanation
    of I .
    (c) Statement I is true , Statement II is false
    (d) Statement I is false , Statement II is true .
    4. Let A and B be two symmetric matrices of the same order . (1)
    Statement I :AB-BA is skew symmetric
    Statement II: AB+BA is symmetric
    Solution 4 . 𝐴′ = 𝐴; 𝐵′ = 𝐵
    Now (𝐴𝐵 − 𝐵𝐴)′ = (𝐴𝐵)′ − (𝐵𝐴)′ = 𝐵′ 𝐴′ − 𝐴′ 𝐵′ = 𝐵𝐴 − 𝐴𝐵
    (𝐴𝐵 − 𝐵𝐴)′ = −(𝐴𝐵 − 𝐵𝐴) ⇒ 𝐴𝐵 − 𝐵𝐴 𝑖𝑠 𝑠𝑘𝑒𝑤 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐
    (𝐴𝐵 + 𝐵𝐴)′ = (𝐴𝐵)′ + (𝐵𝐴)′ = 𝐵′ 𝐴′ + 𝐴′ 𝐵′ = 𝐵𝐴 + 𝐴𝐵
    AB+BA is symmetric
    Option b

    5. Suppose A and B be two square matrices of order 2 (1)

    Statement I : A.adjA=|𝐴|I
    Statement II: adj(AB)=adj(A).adj(B).
    Solution 5. Option (c)
    1 2 3 −7 −8 −9
    6. X. [ ]=[ ] , then find the matrix X
    4 5 6 2 4 6
    𝑎 𝑏
    Solution 6 Let X = [ ]
    𝑐 𝑑
    𝑎 𝑏 1 2 3 −7 −8 −9
    [ ][ ]=[ ]
    𝑐 𝑑 4 5 6 2 4 6
    𝑎 + 4𝑏 2𝑎 + 5𝑏 3𝑎 + 6𝑏 −7 −8 −9
    ⇒[ ]=[ ]
    𝑐 + 4𝑑 2𝑐 + 5𝑑 3𝑐 + 6𝑑 2 4 6
    ⇒ 𝑎 + 4𝑏 = −7,2𝑎 + 5𝑏 = −8,3𝑎 + 6𝑏 = −9
    𝑐 + 4𝑑 = 2,2𝑐 + 5𝑑 = 4,3𝑐 + 6𝑑 = 6
    On solving a=1 , b=-2 , c=2 , d=0
    1 0 2
    7. If A = [0 2 1] and 𝐴3 − 6𝐴2 + 7𝐴 + 𝑘𝐼 = 𝑂, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑘 (3)
    2 0 3

    5 0 8
    Solution 7.𝐴2 = 𝐴. 𝐴 = [2 4 5 ]
    8 0 13
    21 0 34
    𝐴3 = 𝐴2 . 𝐴 = [12 8 23]
    34 0 55
    Given 𝐴3 − 6𝐴2 + 7𝐴 + 𝑘𝐼 = 𝑂
    21 0 34 5 0 8 1 0 2 1 0 0
    [12 8 23] − 6 [2 4 5 ] + 7 [0 2 1] + 𝑘 [0 1 0]=O
    34 0 55 8 0 13 2 0 3 0 0 1
    21 − 30 + 7 + 𝑘 0 34 − 48 + 14 0 0 0
    [ 12 − 12 8 − 24 + 14 + 𝑘 23 − 30 + 7 ] = [ 0 0 0]
    34 − 48 + 14 0 55 − 78 + 21 + 𝑘 0 0 0
    Hence k=2

    8. A diet is to contain 30 units of vitamin A, 40 units of vitamin B and 20 units of


    vitamin C. Three types of foods 𝐹1 , 𝐹2 𝑎𝑛𝑑 𝐹3 are available . One unit of food 𝐹1
    contains 3 units of vitamin A, 2 units of vitamin B and 1 unit of vitamin C . One
    unit of food 𝐹2 contains 1 unit of vitamin A, 2 units of vitamin B and 1 unit of
    vitamin C. One unit of food 𝐹3 contains 5 units of vitamin A , 3 units of vitamin B
    and 2 units of vitamin C .

    Based on above information ,answer the following questions :


    i) If the diet contains x units of food 𝐹1 , y units of food 𝐹2 and z units of
    food 𝐹3 ,then what is the matrix equation representing the above situation
    ?
    ii) If A is the matrix in above situation ,what is the value of |𝑎𝑑𝑗 𝐴| ?
    iii) What is the values of x,y, z in this case ?
    iv) What is the value of |𝐴−1 | ?
    Solution 8. i) 3x+y+5z=30 , 2x+2y+3z=40 , x+y+2z=20
    𝟑 𝟏 𝟓 𝑥 30
    ii) A= [𝟐 𝟐 𝟑] , X=[𝑦] , B= [40]
    𝟏 𝟏 𝟐 𝑧 20
    |𝐴| = 3(4 − 3) − 1(4 − 3) + 5(2 − 2) = 2
    |𝑎𝑑𝑗𝐴| = |𝐴|2 ⇒ |𝑎𝑑𝑗𝐴| = 4
    iii) Above system of equations can be written as AX = B
    |𝐴| = 2 ≠ 0 ⇒ 𝐴−1 𝑒𝑥𝑖𝑠𝑡
    1 3 −7
    Adj A = [−1 1 1]
    0 −2 4
    𝑥 1 3 −7 30 5
    1
    ∴ [ ] = . [−1
    𝑦 1 1 ] [40]=[15]
    2
    𝑧 0 −2 4 20 0
    ⇒ 𝑥 = 5, 𝑦 = 15, 𝑧 = 0
    1 1
    iv ) |𝐴−1 | = |𝐴|−1 = |𝐴| =
    2

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