Matrices and Determinants 4 Question 16

17. The number of value of k, for which the system of equation

(k+1)x+8y=4ykx+(k+3)y=3k1

(2013 Main)

has no solution, is

(a) infinite

(b) 1

(c) 2

(d) 3

Show Answer

Answer:

Correct Answer: 17. (d)

Solution:

  1. Given equations can be written in matrix form AX=B

where, A=k+18 kk+3,X=xy and B=4k3k1

For no solution, |A|=0 and (adjA)B0

Now, |A|=k+18 kk+3=0

(k2+1)(k+3)8k=0k2+4k+38k=0k24k×3=0(k1)(k3)=0k=1,k=3,

 Now adjA=k+38kk+1

Now, (adjA)B=k+384k kk+13k+1

=(k+3)(4k)8(3k1)4k2+(k+1)(3k1)=4k212k+8k2+2k1

Put k=1

(adjA)B=412+81+21=00 not true 

Put k=3

(adjA)B=3636+89+61=480 true 

Hence, required value of k is 3 .

Alternate Solution

Condition for the system of equations has no solution is

a1a2=b1b2c1c2k+1k=8k+34k3k1

Take k+1k=8k+3

k2+4k+3=8k

k24k+3

(k1)(k3)=0

k=1,3

If k1, then 81+34.12, false

And, if k=3, then 864.391, true

Therefore, k=3

Hence, only one value of k exist.

x1



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