Question

plseeeeeeeeeee can any one solve this
i tried this sum but not getting it

Answer 1

No wonder... it's going to have infinitely many solutions. So do as the choices suggest, and set z = k
where k is some fixed constant, and pick any two of the three equations, and solve the new system.

Answer 2

okay

Answer 3

how

Answer 4

i have put k in place of z , till then det = 0 is coming

Answer 5

No, see, k is just a constant. Once you have substituted k for z, pick any TWO of your equations, and solve THAT system.

Answer 6

but using matrices, we have to solve it

Answer 7

Fine... so let's have these equations...
\[\huge \boxed1 \quad 2x+y+z=3\\\huge \boxed2\quad x-2y-z=1\\\huge \boxed3 \quad 3x+4y+3z=5\]

Answer 8

Now, replacing z with k...
\[\huge \boxed1 \quad 2x+y+\color{red}k=3\\\huge \boxed2\quad x-2y-\color{red}k=1\\\huge \boxed3 \quad 3x+4y+3\color{red}k=5\]

Answer 9

Now, pick two of these equations...

Answer 10

simply solve them by elimination method....

Answer 11

i solved it by determinats, and i got my answer

Answer 12

its last one..

Answer 13

Cramer's rule?

Answer 14

yes @terenzreignz

Answer 15

Well, I wouldn't want to let you have all the fun :P
I pick equations 1 and 2.
\[\huge \left|\begin{matrix}2 & 1 \\ 1 & -2 \end{matrix}\right|=-5\]

Answer 16

okay bro
no pblm

Answer 17

Solving for x (replacing the column corresponding to x with the right-hand-side column...
\[\huge \left|\begin{matrix}3-k & 1 \\ 1+k & -2 \end{matrix}\right|=-7+k\]
So.\[\huge x = \frac{-7+k}{-5}=\frac75-\frac{k}5\]


Now, solving for y...
\[\huge \left|\begin{matrix}2 & 3-k \\ 1 &1+k \end{matrix}\right|=-1+3k\]
So. \[\huge y = \frac{-1+3k}{-5}=\frac15-\frac{3}{5}k\]