Solutions of Question 7 and 8 of Exercise 2.6 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
If A=[3214−1273−3]; find A−1 and hence solve the system of equations.
3x+4y+7z=14;2x−y+3z=4;x+2y−3z=0.
Solution.
Given A=[3214−1273−3]|A|=3(−3)−2(−26)+19=−9+52+19=62≠0 This system is consistent. Now to find A−1, we calculate the cofactors of each element . A11=(−1)1+1|−123−3|=3−6=−3A12=(−1)1+2|427−3|=−(−12−14)=26A13=(−1)1+3|4−173|=12+7=19A21=(−1)2+1|213−3|=−(−6−3)=9A22=(−1)2+2|317−3|=−9−7=−16A23=(−1)2+3|3273|=−(9−14)=5A31=(−1)3+1|21−12|=4+1=5A32=(−1)3+2|3142|=−(6−4)=−2A33=(−1)3+3|324−1|=−3−8=−11A=[−326199−1655−2−11]adj(A)=[−39526−16−2195−11]A−1=162[−39526−16−2195−11]A−1=[−3629625622662−1662−2621962−562−1162]A−1=[−3629625621332−832−1321962−562−1162] Therefore, the inverse of matrix A is: A−1=[−3629625621332−832−1321962−562−1162] Now given the system of equations: 3x+4y+7z=142x−y+3z=4x+2y−3z=0 The associated augmented matrix for this system is: Ab=[347142−13412−30]∼R=[12−302−13434714]R1interchangeR3∼R=[12−300−5940−21614]R2−2R1andR3−3R1∼R=[12−300−5940−187]12R3∼R=[10131400−31−310−187]R1+2R3andR2−5R3∼R=[1013140−18700−31−31]R2interchangeR3∼R=[1013140−1870011]−131R2interchangeR3∼R=[100110−10−10011]R2−8R3R1−13R3 From above equation we get, x1=1x2=1x3=1 Now solutions of above equations are; [−3629625621332−832−1321962−562−1162]1;1;1
Determine the value of λ for which the following system has no solution, unique solution or infinitely many solutions.
x+2y−3z=4;3x−y+5z=2;4x+y+(λ2−14)z=λ+2
Solution.