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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 1Given equations can be written in matrix from AX = B where, A=[k+18kk+3], X=[xy] and B=[4k3k−1] For no solution, |A| = 0 and (adj A) B ≠ 0 Now, |A|=[k+18kk+3]=0 ⇒ (k+1)(k+3)−8k=0k2+4k+3−8k=0⇒ k2−4k+3=0⇒ (k−1)(k−3)=0⇒ k=1, k=3, Now adj A=[k+3−8−kk+1] Now, (adj A)B=[k+3−8−kk+1][4k3k+1] =[(k+3)(4k) −8(3k−1)−4k2+(k+1)(3k−1)]=[4k2−12k+8−k2+2k−1] Put k = 1 (adj A)B=[4−12+8−1+2−1]=[00] not true Put k = 3 (adj A)B=[36−36+8−9+ 6−1]=[8−4]≠0 true Hence, required value of k is 3. Alternate Solution Condition for the system of equations has no solution is a1a2=b1b2≠c1c2∴ k+1k=8k+3≠4k3k−1Take k+1k=8k+3⇒ k2+4k+3=8k⇒ k2−4k+3=0⇒ (k−1)(k−3)=0 k = 1, 3 If k = 1, then 81+3≠4.12, false And, if k = 3, then 86≠4.39−1, true Therefore, k = 3 Hence, only one value of k exist.

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