Solve for z only using Cramers Rule 3x y 2z 1 y 5z 0 2x

Solve for z only using Cramer\'s Rule. {3x + y + 2z = 1 -y + 5z = 0 2x + y + 4z = 0

Solution

The determinant of the coefficient matrix of the given system of linear equations is D =

3

1

2

0

-1

5

2

1

4

On computing, D = -13.

The determinant,after replacing the column of coefficients of x by the column vector(1,0,0)^Tis D_x =

1

1

2

0

-1

5

0

1

4

On computing, D_x = -9.

The determinant,after replacing the column of coefficients of y by the column vector(1,0,0)^Tis D_y=

3

1

2

0

0

5

2

0

4

On computing, D_y =10.

The determinant,after replacing the column of coefficients of z by the column vector(1,0,0)^Tis D_z=

3

1

1

0

-1

0

2

1

0

On computing, D_z =2.

Hence x = D_x/D = -9/-13 = 9/13, y = D_y/D =10/-13 = -10/13 and z = D_z/D =2/-13 = -2/13.

We can substitute x = 9/13, y = -10/13 and z= -2/13 in the original equations to verify the result.

The answer is (x,y,z) = (9/13,-10/13,-2/13).

3	1	2
0	-1	5
2	1	4