Question
For what values of k, does the system of linear equation x+y+z=2, 2x+y−z=3, 3x+2y+kz=4 have a unique solution?
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[d] The given system of equations is x+y+z=2 ... (i) 2x+y−z=3 ... (ii) and 3x+2y+kz=4 ... (iii) This system has a unique solution if ∣∣∣∣1231121−1k∣∣∣∣≠0 Applying C2→C2−C1 and C3→C3−C1 We get ∣∣∣∣1330−1−10−3k−3∣∣∣∣≠0 ⇒−1(k−3)−3≠0 or −k+3−3≠0⇒k≠0
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Question Text | For what values of k, does the system of linear equation x+y+z=2, 2x+y−z=3, 3x+2y+kz=4 have a unique solution? |
Answer Type | Text solution:1 |
Upvotes | 150 |