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Question: The value of $k \in R$, for which the following system of linear equations $3x-y+4z=3$ $x+2y-3z=-2$...

The value of k∈Rk \in R, for which the following system of linear equations

3x−y+4z=33x-y+4z=3 x+2y−3z=−2x+2y-3z=-2 6x+5y+kz=−36x+5y+kz= -3

has infinitely many solutions, is

A

Option 1

B

Option 2

C

Option 3

D

Option 4

Answer

-5

Explanation

Solution

We have the system

3x−y+4z=3(1)x+2y−3z=−2(2)6x+5y+kz=−3(3)\begin{aligned} 3x - y + 4z &= 3\quad\quad(1)\\[4mm] x + 2y - 3z &= -2\quad\quad(2)\\[4mm] 6x + 5y + kz &= -3\quad\quad(3) \end{aligned}

Step 1: Express xx and yy in terms of zz using (1) and (2).

From (1), solve for yy:

y=3x+4z−3.y = 3x + 4z - 3.

Substitute into (2):

x+2(3x+4z−3)−3z=−2.x + 2(3x+4z-3) - 3z = -2. x+6x+8z−6−3z=−2⇒7x+5z=4.x + 6x + 8z - 6 - 3z = -2 \quad\Rightarrow\quad 7x + 5z = 4.

Thus,

x=4−5z7.x = \frac{4 - 5z}{7}.

Now substitute xx back into the expression for yy:

y=3(4−5z7)+4z−3=12−15z7+28z−217=−9+13z7.y = 3\left(\frac{4-5z}{7}\right) + 4z - 3 = \frac{12-15z}{7} + \frac{28z-21}{7} = \frac{-9 + 13z}{7}.

Step 2: Substitute xx and yy in (3).

Plug into (3):

6(4−5z7)+5(−9+13z7)+kz=−3.6\left(\frac{4-5z}{7}\right) + 5\left(\frac{-9+13z}{7}\right) + kz = -3.

Simplify:

24−30z−45+65z7+kz=−3,\frac{24 - 30z - 45 + 65z}{7} + kz = -3, −21+35z7+kz=−3.\frac{-21 + 35z}{7} + kz = -3.

Since −217=−3\frac{-21}{7} = -3 and 35z7=5z\frac{35z}{7} = 5z, we have:

−3+5z+kz=−3.-3 + 5z + kz = -3.

Cancelling −3-3 from both sides:

(5+k)z=0.(5+k)z = 0.

For the system to have infinitely many solutions (i.e. hold for all zz), the coefficient of zz must be zero:

5+k=0⇒k=−5.5 + k = 0 \quad\Rightarrow\quad k = -5.