Q. 285.0( 1 Vote )

# Using matrices so

Answer :

Given: system of equations: x+ 2y +z =7, x + 3z =11 and 2x – 3y =1

To find: the solution of the given system of equations

given system of equations is

x + 2y +z =7

x + 3z =11

2x – 3y =1

Now we will write the system of the 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., the determinant of matrix A, so  |A|=1(0-3× (-3))-2(1×0-2× 3)+1(1× (-3)- 2× 0)

|A|=1(0+9)-2(0-6)+1(-3-0)

|A|=1(9)-2(-6)+1(-3)

|A|=9+12-3

|A|=18…………..(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.(9)=9

A12=(-1)1+2.M12=(-1)3.(-6)=6

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

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

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

A23=(-1)2+3.M23=(-1)5.(-7)=7

A31=(-1)3+1.M31=(-1)4.(6)=6

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

A33=(-1)3+3.M33=(-1)4.(-2)=-2

Thus, Now, substituting the above value in equation (iii), we get Substituting value from equation (i) in the above equation, we get Substituting this value in equation (ii), we get Multiplying the two matrices, we get     On equating we get

x=2, y=1, z=3

Hence the solution of the given system of equation using matrices is x=2, y=1, z=3

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