Q. 284.5( 2 Votes )

Using matrices so

Answer :

Given: system of equations: x+y–z=3; 2x+3y+z=10; 3x–y–7z=1


To find: the solution of the given system of equation using matrices


given system of equations is


x+y–z=3


2x+3y+z=10


3x–y–7z=1


Now we will write the system of equation as AX=B,


i.e., in matrix A there will be coefficients of x, y and z,


in matrix X there will be variables x, y and z,


in matrix B will all the constant terms,


so the given system of equations can be written as



So


Now we will solve for |A|, i.e., determinant of matrix A, so




|A|=1(3× (-7)-1× (-1))-1(2× (-7)-1× 3)-1(2× (-1)- 3× 3)


|A|=1(-21+1)-1(-14-3)-1(-2-9)


|A|=1(-20)-1(-17)-1(-11)


|A|=-20+17+11


|A|=8…………..(i)


As |A|≠0, the given system of equation is consistent and has unique solution


Now given system of equation is written as


AX=B


Now the solution of the given system of equation can be calculated as


X=A-1B




And


Now we know


The adjugate, classical adjoint, or adjunct of a square matrix is the transpose of its cofactor matrix.













A11=(-1)1+1.M11=(-1)2.(-20)=-20


A12=(-1)1+2.M12=(-1)3.(-17)=17


A13=(-1)1+3.M13=(-1)4.(-11)=-11


A21=(-1)2+1.M21=(-1)3.(-8)=8


A22=(-1)2+2.M22 =(-1)4.(-4)=-4


A23=(-1)2+3.M23=(-1)5.(-4)=4


A31=(-1)3+1.M31=(-1)4.(4)=4


A32=(-1)3+2.M32=(-1)5.(3)=-3


A33=(-1)3+3.M33=(-1)4.(1)=1


Thus,



Now, substituting the above value in equation (iii), we get



Substituting value from equation (i) in above equation, we get



Substituting this value in equation (ii), we get



Multiplying the two matrices, we get







On equating we get


x=3, y=1, z=1


Hence the solution of the given system of equation using matrices is x=3, y=1, z=1


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