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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 no solution.

The solution is pa and q=b.
Find ba?

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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 No solution D=0 and at least one D1,D2,D3 is not zero, if p=2,D1=D2=D3=D=0
For No solution, p2 and q=3
ba=1

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