# Consider the linear system x+3y+kz=-5 kx+(3+3k)y+(6+k^2)z=-11 x+4y+k^2z=k-4 For the following values of k, k=3 and k=2, write the solutions in vector form.

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Student Comments

pramodpandey | Student

K=3,

x+3y+3z=5 (i)

3x+12y+15z=-11 (ii)

x+4y+9z=-1 (iii)

Multiply (i) by -3 and add to (ii),

3y+6z=-26 (iv)

Subtract (i) from (iii) ,we have

y+z=-6 (v)

Multiply (v) by -3 and add to (iv) ,we have

12z=8

z=2/3

substitute z in (v) ,we have

y=-10

Sustitute y and z in (i) ,we have x=33

when K=3 , (x,y,z)=(33,-10,2/3)

**K=2**

x+2y+2z=5 (i)

2x+9y+10z=-11 (ii)

x+4y+4z=-2 (iii)

Multiply (i) by -2 and add to (ii),

5y+6z=-21 (iv)

Subtract (i) from (iii) ,we have

2y+2z=-7 (v)

Multiply (v) by -2 and add to (iv) ,we have

y=0

substitute y in (v) ,we have

z=-7/2

Sustitute y and z in (i) ,we have x=12

when K=2 , (x,y,z)=(12,0,-7/2)