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Question : 96 of 160
Marks:
+1,
-0
Solution:
Let
Δ=|| x+2 | x+3 | x+5 |
| x+4 | x+6 | x+9 |
| x+8 | x+11 | x+15 |
| Apply operations
R2→R2−R1,R3→R3−R1, we get
Δ=|| Again, apply operation
C2→C2−C1,
C3→C3−C1, we get
Δ=|| Expand along
R1, we get
Δ=(x+2)(4−4)−1(8−12) +3(4−6) =0+4+3(−2)=4−6=−2
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