# Using matrices so

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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