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Question

For what values of p and q, the system of equations
2x+py+6z=8
x+2y+qz=5
x+y+3z=4 has infinitely many solutions.
Enter the value of p

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Solution

Given equations are
2x+py+6z=8
x+2y+qz=5
x+y+3z=4
Here D=∣ ∣2p612q113∣ ∣
Applying R1R12R3
=∣ ∣0p2012q113∣ ∣
=(p2)1q13
=(p2)(3q)
=(p2)(q3)
D1=∣ ∣8p652q413∣ ∣
Applying C1C14C2 & C3C33C2
D1=∣ ∣84pp63p32q6010∣ ∣
Expanding along R3, then
D1=(1)84p63p3q6
=(1){(84p)(q6)+3(63p)}
=4(p2)(q6)+3(p2)3
=(p2)(4q15)
and D2=∣ ∣28615q143∣ ∣
Applying R12R3
=∣ ∣00015q143∣ ∣
=0
and D3=∣ ∣2p8125114∣ ∣
Applying R1R12R3
D3=∣ ∣0p20125114∣ ∣
Expanding along R1 then
D3=(p2)1514
C3=(p2)
By Cramer's rule :

x=D1D,y=D2D,z=D3D
For infinitely many solutions :
D=D1=D2=D3=0, p=2

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