Question

solve by determinants

a.x+y+z=0

3x-y+2z=-1

2x+3y+3z=-5

b. x+2z=1

2x-3y=3

y+z=1

c. x+y+z=10

3x-y=0

3y-2z=-3

d. -8x+5z=-19

-7x+5y=4

-2y+3z=3

e. -x+2y+z-5=0

3x-y-z+7=0

-2x+4y+2z-10=0

f. 1/x+1/y+1/z=12

4/x-3/y=0

2/y-1/z=3

Answer 1

a) Given, x+y+z = 0

3x-y+2z = -1

2x+3y+3z = -5

Here the coefficient matrix is, A = .

Then, D = det(A) = -3

Therefore, there exists a unique a solution for x,y,z.

Now, D_x = = -15

D_y = = -6

D_z = = 21

Therefore, x = (D_x/D) = (-15)/(-3) = 5

y = (D_y/D) = (-6)/(-3) = 2

z = (D_z/D) = (21)/(-3) = -7

Hence, the solution is x = 5, y = 2, z = -7.

b) Given, x+2z = 1

2x-3y = 3

y+z = 1

Here the coefficient matrix is A = .

Now, D = det(A) = 1

Therefore, there exists a unique solution for x,y,z.

By Cramer's rule, x = = 9/1 = 9

y = = 5/1 = 5

z = = (-4)/1 = -4

Hence, the solution is x = 9, y = 5, z = -4.

c) Given, x+y+z = 10

3x-y = 0

3y-2z = -3

Here, the coeeficient matrix is A = .

Now, D = det(A) = 17

Therefore, there exists a unique solution for x,y,z.

By Cramer's rule, x = = 17/17 = 1

y = = 51/17 = 3

z = = 102/17 = 6

Hence, the solution is x = 1, y = 3, z = 6.

d) Given, -8x+5z = -19

-7x+5y = 4

-2y+3z = 3

Here the coefficient matrix is A = .

Now, D = det(A) = -50.

Therefore, there exists a unique solution for x,y,z.

By Cramer's rule, x = = (-400)/(-50) = 8

y = = (-600)/(-50) = 12

z = = (-450)/(-50) = 9

Hence, the solution is x = 8, y = 12, z = 9.

e) Given, -x+2y+z = 5

3x-y-z = -7

-2x+4y+2z = 10

Here the coefficient matrix is A = .

Now, D = det(A) = 0.

Therefore, there exists infifnitely many solutions of the system.

f) Given, (1/x)+(1/y)+(1/z) = 12

(4/x)-(3/y) = 0

(2/y)-(1/z) = 3

Here the coefficient matrix is A =

Now, D = det(A) = 15

Therefore, there exists a unique solution for x,y,z.

By Cramer's rule, (1/x) = = 45/15 = 3

(1/y) = = 60/15 = 4

(1/z) = = 75/15 = 5

Hence, the solution is 1/x = 3, 1/y = 4, 1/z = 5 ,i.e., x = 1/3, y = 1/4, z = 1/5.