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Question

The number of value of k, for which the system the system of equation (k + 1)x + 8y = 4k kx + (k + 3)y = 3k - 1 has no solution, is

A
infinite
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B
1
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C
2
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D
3
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Solution

The correct option is B 1
Given equations can be written in matrix from AX = B
where, A=[k+18kk+3], X=[xy] and B=[4k3k1]
For no solution, |A| = 0 and (adj A) B 0
Now, |A|=[k+18kk+3]=0
(k+1)(k+3)8k=0k2+4k+38k=0 k24k+3=0 (k1)(k3)=0 k=1, k=3,
Now adj A=[k+38kk+1]
Now, (adj A)B=[k+38kk+1][4k3k+1]
=[(k+3)(4k) 8(3k1)4k2+(k+1)(3k1)]=[4k212k+8k2+2k1]
Put k = 1
(adj A)B=[412+81+21]=[00] not true
Put k = 3
(adj A)B=[3636+89+ 61]=[84]0 true
Hence, required value of k is 3.
Alternate Solution
Condition for the system of equations has no solution is
a1a2=b1b2c1c2 k+1k=8k+34k3k1Take k+1k=8k+3 k2+4k+3=8k k24k+3=0 (k1)(k3)=0
k = 1, 3
If k = 1, then 81+34.12, false
And, if k = 3, then 864.391, true
Therefore, k = 3
Hence, only one value of k exist.

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