Question

Problem 3. Solve the value of x, y, and z of the given system of equations using matrix algebra. (1) 6x + 8y -7z=-145 9x-3y -62 = -180 -5x + 12y + 4z = 98

Answer 1

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Solutions Given System of equations is given as- 65+ 8y-72 = -145 98-34-62 = -180 -52+12y +4z = 96 In matrix form above system is written as AX=B |--145 6 0 -7 o where, A- x= B= 9 -3 as -180 -6 4 -5 12 Z 98 multiplying by À to ean by the left side- A'AX AB A=I 3) I X=AB =) X=AB 2 is the solution. where, Adjoint of A 3 det(A) T -Cl Now, Adjoint of A= (12 C13 C22 C23 [call C21 C31 C32 C33 where Cij cofactor of element of all 1 Page Laf 3

T Adjoint (A)- +(-12+72) -136-30) +(108-15) -(32+84) +124-35) -(72+40) [+1-48-21) -(-36+63) +(-18-72) T 6o - 6 93 y Adjoint (A)=-116 -112 - 27 -go -69 T60 - 16 -69 2 Adjoint (A)= -6 -11 -27 93 -112 -go det(A)= 66-12 +72) –0 (36-30) -7(108-15) = 6 (60) – 8(6) -7193) de t(A) = 360 48 - 651 = -339 Now from ③ 60 -116 -69 Ta -27 11 -6 339 93 -112 -go from @ - 116 -69 5:45 60 -6 93 -11 -27 -180 X 339 -112 -90 98 Page 293

X 339 60X(-145) - 116X(-180) - 6998 -6X6-145) -11 X (-180) -27 X98 93x4-145) -112x6-180) - 90 x 98 => x= -1 339 -8700 +20080 – 6762 8 70 + 1980 – 2646 |-13485 +20160 2020 » X= -1 5418 339 204 |- 2145 5418/339 [1806/113 ra al y -204/339 - 68/113 715/113 2145/339 solution is -1806 113 2= - 68 ya 113 715 z = 113 page 383